





Book_ __i.1 _Li 

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engineers’ manual. 


I 


Chas. C. Moore & Co. 

AGENTS FOR 

BABCOCK & WILCOX BOILERS 

McIntosh and Seymour Engines, 

Hamilton Corliss Engines, 

N. Y. Safety Automatic Engines, 

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Hoppe Feedwater Purifiers, 

Goubert Feedwater Heaters. 

Stratton Steam Separators, 

Snow Pumps, 

Quimby Screw Pumps, 

Chapman Valves, 

Spencer Damper Regulators, 

Giepel Steam Traps, 

Hyatt Roller Bearings. 

SEND FOR CATALOGUES. 

BRANCH OFFICES: 

218 SOUTH SECOND AVE., SEATTLE, WASH.. 
103 SOUTH BROADWAY, LOS ANGELES, CAL. 

MAIN OFFICE: 

32 FIRST ST., SAN FRANCISCO, CAL. 


II 


ENGINEERS* MANUAL. 


WHEN IN NEED OF ENGINEERS’ SUPPLIES 
OF ANY KIND CALL AND SEE, OR PHONE 


Niki, Lewis {Staner Co. 


MINING, MILL, MARINE AND FARM 

Machinery and 

Supplies «*■>**-»> 


LONG DISTANCE PHONE MAIN 89. 


308-310 FIRST AYE. SOUTH, SEATTLE, WASH. 


OILS 


PACKING 






THE 

Seattle-Pacific Coast Engineers’ 
Electricians’ Machinists’ 


MANUAL AND DIRECTORY 



SEATTLE UNION 

LOCAL NO. 40 

SEATTTE, WASHINGTON 


Copyrighted by 

wm. u. McCarthy 


»* » J } 

<s> ) .) 

> > p 0 it 


J'J) 
) ) > 


Price One Dollar 



PRESS, SEATTLE 





THE METROPOLITAN 




6 


THE SR ARY OF 

CONGRESS, 
Two Gooses Receiveo 

NOV. 30 1901 

COPYRIGHT ENTRY 

CLASS CuKXc. Ho. 

%>■ i ^ 7 

COPY -j. 


ENGINEERS’ MANUAL 


A FEW 




of the Articles we Carry in Stock 


Boilers, Engines, Shingle Machines, Saw Mills, Hoisting 
Engines, Steam Pumps, Air Compressors, Aair and Circulat¬ 
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Pumps, Boiler Feed Pumps, Shafting, Pulleys, Hangers, Boxes, 
Set Collars, Belting, Packing, Lace Leather, Hose, Sheet Rub¬ 
ber, Cap Screws, Studs, Machine Bolts, Rough Nuts, Semi-fin¬ 
ished Nuts, Gas Pipe, Water Pipe, Valves of all kinds, Emery 
Wheels, Files, Oil Cans, Injectors, Inspirators, Lubricators, 
Sheet Copper, Sheet Brass, Copper Rod, Copper Pipe, Brass 
Pipe, Brass Rod, Lathe Chucks, Lathe Dogs, Lathe Files, Ma- 
chnists’ Vises, Pipe Vsies. 

In addition to our store, we have the best equipped ma¬ 
chine shop in the state of Washington. We mean by being 
the best equipped machine shop that our tools are the most 
« modern,, having- «thp *bG$t range of work, and are the newest 
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Corner Washington Street and Railroad Avenue, Seattle. 

- 









ENGINEERS’ manual. 


7 


PLAIN FACTS FOR ENGINEERS. 


Why You Should Become a Member of the International 
Union of Steam Engineers. 


1st. Because it develops fraternity. Craftsmen are all too 
jealous and suspicious of one another, even at the best. 

2d. Because it gives men self-reliance and confidence in 
one another. A servile boss truckler is not a free man. 

3d. Because it is a good investment. No other investment 
gives back so large a return for expenditure of time and 
money. 

4th. Because it educates as to public questions. Our Union 
takes the place of the debating club and professor’s lecture. 

5th. Because it impels a man, and enables him to gain the 
best possible kind of a livelihood for himself and those who 
are dependent upon his efforts. 

6th. Because the engineer, notwithstanding the responsible 
position he holds with lives and property in his care, as a 
general thing works longer hours, harder, and is the poorest 
paid mechanic in the country. 

7th. Because in union there is strength. This is as true 
of engineers as of states. Labor unions stand as a fortress 
for the defense of labor. 

8th. Because skinflints and dilletanti condemn it. The 
trade union is to be recommended for the enemies it has 
made. The non-union man is the sutler of the unon army. 

9th. Because we believe engineers should be organized for 
themelsves. Agents, drummers and others have no business 
in engineers’ organizations. One is liable to make a tool of 
the other. 




/ 


8 


ENGINEERS’ MANUAL. 


J. A. SODERBERG, Pres, and Treas. 

C. A. PACKARD, Vice-Pres. and Mgr. 

C. J. CARLSON, Secretary. 



Soderberg Pipe 



(INCORPORATED.) 


Manufacturers of. 


.Heme Wood Pipe 


INSTALLERS OF WATER SYSTEMS, DREDGE 
PIPE, IRRIGATION PIPE, STEAM AND 
HEAT INSULATION, ELECTRIC 
WIRE, OIL AND SALT 
WATER CONDUITS. 

OFFICE: 410-11 WASHINGTON BLDG. 
Telephone Red 301. 

SEATTLE, WASHINGTON. 














engineers’ manual. 


9 


DECLARATION OF OUR PRINCIPLES. 

/ . ■' - 

Trade Unions are tlie bulwark of civilization. 

— Wm. E. Gladstone. 

Resolved, That we, as a body, thoroughly approve of the objects 
of the American Federation of Labor, and pledge ourselves to give 
it our earnest and hearty support. 

Resolved, That the members of this organization should make 
it a rule, when purchasing goods, to call for those which bear the 
trade-mark of organized labor, and when any individual, firm or 
corporation shall strike a blow at labor organizations, they are 
earnestly requested to give that individual, firm or corporation their 
careful consideration. No good unipn man can kiss the rod that 
whips him. 

Resolved, That it is of importance that members should vote 
intelligently, hence the members of this organization shall strive to 
secure legislation in favor of those who produce the wealth of the 

country. 

« 

Resolved. That we hold it as a sacred principle that the trade 
union men, above all others, should set a good example as good and 
faithful workmen, performing their duties to their employers with 
honor to themselves and their organizations. We hold that a reduc¬ 
tion of hours for a day’s work increases the intelligence and happi¬ 
ness of the laborer, and also increases the demand for labor and the 
price of a day’s work. 

We recognize that the interests of all classes of labor are identi¬ 
cal, regardless of occupation, nationality or religion; for a wrong- 
done to one is a wrong done to all. 

We object to prison contract labor, because it puts the criminal 
in competition with honest labor, for the purpose of cutting down 
wages, and also because it helps to overstock the market. 

We favor the adoption of the first Monday in September as 
Labor’s holiday, and recommend that all local unions endeavor to 
observe the same. 

Therefore, we pledge ourselves unitedly in behalf of the prin¬ 
ciples herein set forth, to perpetuate our order on the basis of friend¬ 
ship and justice, to expound its objects and work for its general 
adoption, to respect and obey laws laid down for its guidance and 
government. 

Organization is life .—Edmund Burke. 


10 


ENGINEERS’ manual. 



MARSH 

STPHLA.M F= 3 L_nVlF 3 S 

BEST FOR EVERY PURPOSE. 

WILL NOT RACE, RUN AWAY OR POUND 


Largest Stock in the Northwest. 


Simonds Manufacturing Co. 

119 Jackson Street, Seattle, Washington 
SOLE AGENTS 





ARITHMETIC 


Only such rules which are apt to be easily forgotten will be given 
here, and it is understood that some training in Arithmetic and 
Algebra has been previously obtained by the reader. 

To Find the Greatest Common Measure (G. C. M.) or Highest 
Common Factor— 

Rule : Divide the greater by the less; and with the remainder 
divide the divisor and so on until there is no remainder, and the last 
divisor is the G. C. M. 

Example—Find the G.C.M. of 689 and 1,573. Ans. 13. 

To Find the Least Common Multiple of two or more numbers— 

Rule : Divide the given numbers by any number that will divide 
the greatest number without a remainder, and set the quotients with 
the undivided numbers in a line beneath. Divide the second line as 
before and so on until there are no two numbers that can be divided; 
then the product of all the divisors and last quotients will give the 
multiple required. 

Example—Find the L.C.M. of 4, 18, 36, 72: 


4 

4 18 

36 

72 

9 

1 18 

9 

18 

2 

1 2 

1 

2 


1 1 1 1 


L.C.M. = 4 x 9 x 2 = 72. 

Example—What is the L C.M. of 20, 36, 48, 100. Ans. 3,600. 


FRACTIONS. 

To reduce a Fraction to its lowest terms — 

Rule : Divide both numerator and denominator by the G.C.M. 

Example $&|=H=s* 

To change an Improper Fraction to a Mixed Number— 

Rule : Divide numerator by the denominator and the remainder 
placed over the denominator is the fraction, viz., V’ = 9$« 

To change a Mixed Number to an Improper Fraction — 

Rule : Multiply the whole number by the denominator of the 
fraction and to the product add the numerator ; place the sum over 
the denominator, viz., 6£ = ^-. 






12 


ENGINEERS’ manual. 


ALL PATENT DRAWINGS GUARANTEED 

BLUE PRINTING. 


Frank E, Adams 

Solicitor of U. S. and Foreign 

PATENTS 

Expert in Patent Office and Mechanical Drawing. 


io COLMAN BLDG., SEATTLE, WASH. 
PHONE BLUE 570. 

SEARCHES, CAVEATS, DESIGN PATENTS, 
TRADE MARKS. 

IDEAS DEVELOPED, MACHINES DESIGNED, 
WORKING DRAWINGS. 

Maps, Plats and Tracings. 
APPOINTMENTS FOR EVENING CONSULTA¬ 
TION AT RESIDENCE. 

RES. PHONE BLUE 356. 


Branch Offices: Washington, D. C 

OTTAWA, CANADA. 

TACOMA, WASH. 





engineers’ manual. 


13 


To Reduce a Compound to a Simple Fraction— 

Rule : Multiply the numerators together for n new numerator, 
and the denominators together for a new denominator, and then 
reduce to its lowest terms. 

Example r °f ity °f — A* 

To Divide Fractions— 


Rule : Reduce to the form of simple fractions, invert divisor and 
iroceed as in multiplication. 

Example —fpf if | of \ ' U _ 21 s 

i* 3i ~ i * V- ~ ~ 


To Add Fractions— 

Rule : Reduce all to a common denominator, then add the numer¬ 
ators, and place the sum over the common denominator. 

, 12 + 15 + 9 

Add i+t+vv = -—^ ^n=Jb 

To Subtract Fractions— 

Rule : Reduce them to a common denominator, subtract the 
numerators and place the difference over the common denominator 


To Add Decimals— 


Rule : Set down the figures so that the decimal points are one 
above the other, then proceed as in addition. 

12.6798 

.0346 

1.32 

•0035 

5-362I 


19.4000 

To Subtract Decimals— 

Rule : Set down the figures so that the decimal points are above 
one another, and then proceed as in simple subtraction. 

12.7896 

6.6794 


6.1102 

To Multiply Decimals — 

Rule : Proceed as in simple multiplication, then point off as 
many decimal places as there are in the multiplier and multiplicand. 

2.03 

.76 


1218 

1421 


1.5428 








14 


ENGINEERS’ MANUAL. 


I 

i 

I 
I 

i 

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> ♦> ♦> 


TELEPHONE, JAMES 2011. 


ESTIMATES FURNISHED 

AND PLANTS INSTALLED. 


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and 

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BOUGHT AND SOLD. 










engineers’ manual. 


15 


To Divide Decimals— 

Rule : Divide as in whole numbers and point off in the quotient 
as many decimal places as those in the dividend exceed those in the 
divisor. Ciphers must be added to the dividend to make its decimal 
places at least equal to those in the divisor, and as many more as it is 
desired to have in the quotient. 

Example, 33-^.055 = ^~ = 6 oo, or .33-7-55=.006. 

To Convert a Decimal into a Vulgar Fraction— 

Rule : Put down the decimal as the numerator and place as the 
denominator 1 with as many ciphers as there are decimal places in 
the numerator. 

•75 — iVs — ¥• 

To Convert a Vulgar into a Decimal Fraction— 

Rule : Divide the numerator by the denominator, adding as 
many ciphers prefixed by a decimal point as are necessary to give the 
number of decimal places desired in the result. 

i= 1.00 4 = .25. 

To Reduce a Repeating Decimal to a Vulgar Fraction— 

Rule : Subtract the decimal figures that do not repeat from the 
whole decimal, including one set of repeating figures ; set down the 
remainder as the numerator of the fraction, and as many nines as 
there are repeating figures followed by as many ciphers as there are 
non-repeating figures in the denominator. 

Example, .633 = 633 

6 

•57 = tL— 

ALGEBRA. 

Algebra is the science which teaches the use of symbols to denote 
numbers and the operations to which the numbers may be subject. 

Example—Add to a the sum of b and c 

a + (b + c). Ans. a + b + c. 

Subtract the number b from a. Ans. a - b. 

(I.) When a bracket is preceded by the sign + remove the 
bracket and leave the terms unaltered. 

(II.) When a bracket is preceded by the sign - remove the 
bracket and change the sign of each term in it. 

Thus a + b+ (c-d+e — b + c- d+e-f 
and a + b - (c-d+e-f) =a + b - c+ d-e+f 

(I.) In addition attach the lower line to the upper with the signs 
of both lines unchanged. 



16 





ENGINEERS’ MANUAL. 


If you want the BEST get 

MACHINE FILES 
HACK SAWS 
METAL SAWS 




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CORUNDUM WHEELS 
EMERY WHEELS 



Sold by 

California Saw Works 

MAIN AND OCCIDENTAL AVES. 
SEATTLE, WASHINGTON 



PULLEYS. 


BELTING. 




engineers’ manual. 


l 7 

(IT.) In subtraction attach the lower to the upper line with the 
signs of the lower line changed. 

Example, (i) To a+ 6+ 7 
add a — b — 5 
2 a +2 

(2) From a + b + j 
a- b - 6 
2 b+ 13 

The methods of denoting brackets are various. Thus, besides 
the marks ( the marks J or j J- are often used. Sometimes 
the “vinc ulum” is drawn over the symbols which are to be connected, 
thus: a-b + c is used to represent the same expression as a-(b + c), 
In removing brackets from an expression, commence with the inner¬ 
most and remove them one by one, and the outermost last of all. 

Thus, a - [b+ {c- (d- £+/')}] 

= a- [b + {c-\d-e-f)}] 

— cl — [6+{c— 

= a-[b+ c- d-\-e+f\ 

— a- b- c+ d-e-f 

or 5 at-(3^-7) - {4 - 2at- (6at- 3)} 

= IO.V. 


Multiplication—When the factors multiplied have like signs, pre¬ 
fix + and when unlike - to the product. 


Multiply a-\-b by a-b; and a-b by a-b 
a+b a-b 

a^b a-b 


a 2 + cib 
- ab-b 3 
a' 2 - 6 2 


a 2 - ab 
- ab-\-b 2 
a 2 - 2 ab + b 1 * 


Involution.—This is the operation of multiplying a quantity by 
itself any number of times. 

a 2 is called the second power of a. 
a 3 is “ “ third “ “ a. 

The signs of even powers of a negative quantity will be positive 
and of the odd powers negative. 

(-a)*={-a)(-a)=a ‘ 2 
(- a) 3 =(-a)(-a)(-a)=-a\ 

To raise a simple quantity to any power. Multiply the index of 
the quantity by the number denoting the power to which it is to be 
raised and prefix the proper sign. 

Thus the square of a 3 is a 3 
“ cube of a 3 is a 9 
“ “ of — x^yz 3 is -x 6 y 3 z 9 . 

(a + &) 2 =a- 4 2ab+ b 2 









18 


engineers’ manual. 


TELEPHONE RED 3956. 

417 NEW YORK BLOCK, SEATTLE. 


FRANKE L. PARKER 

ENGINEER 


PLANS AND SPECIFICATIONS: 

MILLS 

FACTORIES 

POWER PLANTS 

MINING INSTALLATIONS 

HEATING AND VENTILATION 

SURVEYS AND LEVELS: 

CITY LOTS 
ADDITIONS 
TOWN SITES 
MINING CLAIMS 
ROADS, TRAMWAYS 










engineers’ manual. 


*9 

or the square of the sum of two numbers is equal to the sum of their 
squares and twice their product. 

(«+ b) 3 = a 3 +3 a 2 b + ^ab 2 + b 3 

\a + b) 4 = a 4 + ^a 3 b+ 6a 2 b 2 + 4 ab 3 '+ b 4 

which show that the indices of a decrease by unity in each term and 
that the indices of b increase by unity in each and the numerical 
coefficient of the 2nd term is always the same as the index of the 
power to which the binomial is raised. 

(a - b ) 2 = a 2 - 2ab + b 2 

or the square of the difference of two numbers is equal to the sum of 
their squares—twice their product. 

(a - b)(a + b) — a 2 b 2 . or the product of the sum and difference of 
two numbers is equal to the difference of their squares. 

Division.—When the dividend and the divisor have the same sign 
the quotient is positive, and when they have different signs the quotient 
is negative. 

The following will show the process in easy examples : 

Divide x 6 -y 6 by x 2 - y 2 

x 1 -y 2 J * 6 -y 3 lx 4 +x 2 y 2 +y 4 
x 6 - x 4 y 2 
x 4 y 2 - y 6 
x 4 y 2 - x % y 4 
x 2 y 4 -y 6 

x 2 y 4 -ye 

Divide x 6 - 4 a 2 x 4 + 4 a 4 x 2 - a 6 by x 2 - a 2 

x 2 - a 2 J x 6 - 4a 2 ^ 4 + 4« 4 ;*: 2 - a 6 La: 4 - 3a 2 x 2 + a 4 
x 6 - a 2 x 4 

- 3a 2 x 4 + 4 a 4 x 2 - a 6 

- 3« 2 at 4 + 3« 4 a: 2 

a 4 x 2 — a 6 
a 4 x 2 - a 6 

Simple Equations— 

An equation is a statement that two expressions are equal. A 
simple equation is one which contains the first power only of an un¬ 
known quantity. 

Any term of an equation may be transferred from one side to the 
other if its sign is changed. 

Example, 5^ -8 = 3X+2. 

Transposing the terms we get 5^- ^x=2 + 8. 

Combining like terms, 2x— 10. 

And dividing both sides by 2 we get x=§. 

In a company of 266 persons, composed of men, women, and 
children, there are twice as many men as there are women, and twice 







20 


ENGINEERS’ MANUAL 


*♦* *$♦ *4* ♦£♦ *4* *$♦ ♦** >}♦ *4+ «$» *}* ♦♦♦ +*» ♦$*”*$♦ *J» 

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engineers’ manual. 


21 


as many women as children. How many are there of each ? 

Let x — number of children ; 

2x — number of women ; 

4.x = number of men. 
qx+ 2 x + x= 266 . 
jx— 266. 

x — 38 children ; s 

76 women ; 

152 men. 

Example—A vessel can be filled in 15 minutes by 3 pipes, one of 
which lets in 10 gallons more and the other 4 gallons less than the 
third each minute. The cistern holds 2400 gallons. How much comes 
through each pipe in a minute? 

Ans.: 1st pipe 51^3 gallons per minute ; 

2nd pipe 6r^ gallons per minute ; 

3rd pipe 47/| gallons per minute. 

When several unknown quantities are to be determined, there 
must be as many independent equations as there are unknown quan¬ 
tities. 

Thus a + 6 = 6, from which we could not determine the definite 
value of a and b. We must have a second equation independent of 
the first, then find a pair of values of a and b which will satisfy both 
equations. 

If we give a — b=2 we can find the values 
a + b — 6 
a-b — 2 

By addition 2« = 8 
« = 4 

And by subtraction we get a + b= 6 

a - b — 2 
26=4 
b= 2 

Example, 3X + 7y = 67 

5 ^ + 4^=58 

Multiply first equation by 5 and the second by 3. 

15 ^ + 35 ^ = 335 
15.# + 12 y— 174 

Subtracting 23^ =161 

y= 7 

and since 5 x + 4y= 58 

and substituting the value of y from above we get: 

3X + 2S— 58 

5 * = 3 <> 
x— 6 

If there are three unknown quantities their values may be found 
by three independent equations. For from two of the equations a third 
which involves only two unknown symbols may be found, and from 
the remaining equation, and one of the others, a fourth containing 
only the same two unknown symbols may be found. 


22 


engineers’ manual. 



Successors to 
T. F. CLARK & CO. 



Dust Collectors, Blow Piping 


Relief Switches, Smoke Stacks, Boiler Breeching and 
General Sheet Metal Workers. 


OFFICE 107 W. MADISON. 
TELEPHONE MAIN 487. 


MILL WORK A SPECIALTY. PLANS AND ES¬ 
TIMATES FURNISHED ON APPLI¬ 
CATION. 


SEATTLE, WASH. 







engineers’ manual 


2 3 


Quadratic Equations. 

A quadratic equation is one into which the square of an unknown 
symbol enters with or without the first power of the symbol. 

Thus x- — 9 

and x 2 + $x = 24 

are quadratic equations. 
x 2 =9 is a pure quadratic equation. 
x 2 + 5-r= 24 is an adfected quadratic equation. 

Every pure quadratic equation has two roots equal in magnitude 
but with different signs. 

x 2 = 16 

x =±4 

Adfected Quadratic Equations are solved by adding certain terms 
to both sides of the equation so as to make the left hand a perfect 
square. 

Thus x 2 +6^ = 72 

By adding 9 to each side we get 

x 2 + 6 x + 9 = 72 + 9 

Extracting square root we have x+ 3= ±9 

x =6 or - 12 

Example—A ladder, whose foot rests in a given position, just 
reaches a window on one side of a street, and when turned about its 
foot just reaches a window on the other side. If the two positions 
of the ladder be at right angles to each other and the heights of the 
windows be 36 and 27 feet respectively, find the width of the street 
and the length of the ladder. 

Arithmetical Progression. 


Arithmetical Progression is a series of numbers which increase 
or decrease by a constant difference. 

Thus, 2, 4, 6, 8, 10 

9, 7> 5,3, 1, are arithmetical progressions. 

Let a = first term. 

d— difference. 
n — number of terms. 
s — sum of terms. 
x — last term. 

To Find the Last Term— 

Formula, x — a + (n-i)d 

To Find the Sum— 

n 

Formula, .?=-(a + ^) 

2 


n 

= -{2a + (n - 1 )d } 
2 


To Find the Number of Terms— 


Formula, n~ 


2s x - a + i 
a + x ~ d 



24 


engineers’ manual. 


Met & Richards 


4 * 


* 

4 * 




Dealers in 


Oils, 


■b 




108 FIRST AVENUE SOUTH, 


4 » 


SEATTLE, 


WASH. 








engineers’ manual. 


25 


To Find the First Term— 


Formula, a — x-(n-i)d= 


2 s 


Example—Find the last term of the series, also the sum, 

7, 10, 13, to 20 terms. 

Ans. 64, 710. 

Geometrical Progression. 

Geometrical Progression is a series of numbers which increase or 
decrease by a constant factor. 

3, 6, 12, 24, 48 

16, 4, 1, T \ are Geometrical Progressions. 

The constant factor is usually called the Common Ratio. 

Let a = first term. 

/-common ratio. 
n = number of terms. 

5=sum of the n terms. 
at— last term. 

To Find the Last Term— 

Formula, x^af 1 ^ 1 

To Find the Sum— 

_ , a( f n - 1) f x - a 

/ - 1 /" 1 

To Find the First Term— 

SC 

Formula, a=—~—=fx-(f~i)s 
To Find the Common Factor— 

n i 

a V x 


Formula, f = 


s - x 


Example—Insert 3 Geometric means between 1 and 16. 

From this we get s — 5, and the common factor, 

ra-l 4 __ 

- = 2 
* I 


/= 


Ans. 1, 2, 4, 8, 16. 

Example—Find the sum of 1, 3, 9,. to 6 terms. 

Ans. 364. 

Evolution. 

Evolution is the operation of finding any root of a given number. 
In involution the base and the exponents are given and the power is 
determined therefrom. In evolution the base is to be determined, 








26 


ENGINEERS’ manual. 


ESTABLISHED 1887, 


Shop Phone Main 1127 
Res. Phone, White 441 


Commercial Street 
Boiler Works 

H. W. MARKEY, Proprietor. 


cManufadurer and Repairer of 

c BOILE c R_S 

STATIONARY AND 
MARINE WORK A SPECIALTY. 


cAU Kinds of Sheet Iron Work* 


First Avenue South, 


SEATTLE, WASH. 




ENGINEERS’ manual. 


27 


the power itself being given and also the exponent or index of its 

degree. By prefixing the symbol y/ denotes the square root of the 
'3 6 

given number; y/ denotes the cube root; yf denotes the 5th root. 

Fractional exponents are also used to denote the roots of the numbers 

1 i 1 _3_5_ 

as 81 , 64^, 32 s , which is the same as y/ 8 1 , ^/ 6 ‘ 4 , \/ 32 * 

The 4th root is the square root of the square root. 

The 6th root is the square root of the cube root. 

The 9th root is the cube root of the cube root. 

To extract the square root of 123456.789, commence at the deci¬ 
mal point and mark off the given number into periods of two places 
each in the two directions and add as many ciphers as may be neces¬ 
sary, as 123456.789000 

Find the greatest number whose square is less than the first left 
hand period, and place it as the first figure in the quotient. Subtract 
its square from the left hand period and to the remainder bring down 
the two figures of the second period. Double the first figure of the 
quotient for part of the next divisor ; ascertain how many times the 
latter is contained in the dividend exclusive of the right hand figures, 
and set the figure representing that number of times as the second 
figure in the quotient, and annex to the right of the partial divisor 
forming now the complete divisor. Multiply the divisor by the second 
figure in the quotient, and subtract the product from the dividend. 
To the remainder bring down the next period and proceed as before, 
in each case doubling the figures in the root to obtain the trial divisor. 

123456.789000 (^35^36418 
9 


65 

334 

325 

701 

95 6 

701 

7023 

25578 

21069 

70266 

450990 

421596 

702724 

2939400 

2810896 

7027281 

12850400 

7027281 

70272828 

582311900 

562182604 


The square root of 123456.789 is 351.36418 which can readily be 
proved by squaring this number. 










28 


engineers’ manual. 


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engineers’ manual. 


29 


To extract the square root of a vulgar fraction extract the square 
root of numerator and denominator = f = §, or convert the vulgar 

into a decimal fraction and extract the root y/l = ^.4444 = .6666 or 

To extract the square root of large numbers it is easier done by 
logarithms. 

Example, v/107506 

Log. of 107506=5.03143270 
Divide by 2 = 2.51571635 
Log. 2.51571635 = 327.88= square root. 

Practical Geometry. 

To divide a given triangle ABC into any number of equal parts 
by lines parallel to A B : 

Divide B C into the required number of parts ; upon B C describe 
a semicircle, raise perpendiculars from the points of division, meeting 
the semicircle, with C as centre ; describe arcs from the points of 
intersection of the perpendiculars and the semicircle cutting B C in 1, 
2, 3, 4, etc. Draw parallels to A B from f, 2, 3, 4. 

To divide a triangle into any number of equal parts through the 
apex : 

Divide the base into the required number of parts and join the 
points of division to the apex. The triangles thus formed have equal 
bases and equal altitudes, therefore their areas are equal. 

To bisect any irregular figure by a line drawn from one of its 
corners: 

Let A B C D be the given figure, and A the given corner. Draw 
the diagonals A C B D. Bisect B D in F, and through the point iq 
draw F G cutting B C in G. Join A G and the figure is bisected. 

To divide a square into any number of equal parts by lines drawn 
through one of its corners : 

Let A B CO be the required square. Divide the side B C into 
the required number of parts (say 5), marking the points 1, 2, 3, 4, 5, 
and do the same with the side CD marking the points 6, 7, 8, 9, the 
mark 5 will be at the corner C. Join 2, 4, 6 , 8, to the corner A and 
the figure will be divided as required. 

Mensuration of Surfaces. 

To Find the Area of a Triangle — 

Case I. When base and perpendicular are given. 

Rule : Multiply the base by the perpendicular and divide by 2. 

A base x perpendicular 

Arcs. —-— 

2 

and by transposition we get 

g _ 2 Area 

Perpendicular 

„ j. , 2 Area 

Perpendicular = - 







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engineers’ manual. 


3 1 


Case II. When the three sides are given. 

Rule: From half the sum of the three sides subtract each side 
separately ; multiply the half sum and the three remainders together, 
and extract the square root of the product. 

Let A, B , C, represent the three sides of the triangle, and 
A + B+C 

-= half the sum =5 

2 

then the formula is 

J S{S-A){S-B)(S-C) = Area. 

Example.—What is the area of a triangle whose sides are respec¬ 
tively 9, 10, 12 feet? Ans.—44.03 square feet. 

To Find the Area of a Trapezoid— 

Rule : Multiply half the sum of the two parallel sides by their 
perpendicular distance. 

A trapezoid is a quadrilateral figure with only one pair of 
opposite sides parallel. 

Example—A board 8" wide has its two parallel sides i'-6"and 
2' ~ 3", what fraction of a square yard will it cover ? Ans.—of a 
square yard. 

To Find the Area of a Trapezium. 

Case I. When a diagonal and two perpendiculars are given. 
Rule I. Find the area of each triangle and take the sum. 

Rule II. Multiply half the diagonal by the sum of the per¬ 
pendiculars. 

A trapezium is a quadrilateral figure with unequal sides. 

Case II. When the diagonal and the four sides are given. 

Rule : Find the area of each triangle and take the sum. 
Example.—In a trapezium the diagonal is 80 yards, and the two 
perpendiculars are 36 and 42 yards. What is the area? Ans.—3120 
square yards. 

In a trapezium A B CD, the side A D is 18', D C 14', CB 15', and 
AB 12'; the diagonal A C is 18'. Find the area? Ans.—205.37 
square feet. 

To find the Area of a Parallelogram. 

Rule : Multiply the length by the perpendicular breadth. 
Formula, L.B—A. 

The varieties of parallelograms are the 
Square, having 4 sides equal and all angles right angles. 

Rectangle, having opposite sides equal and all angles right angles. 
Rhombus, having all 4 sides equal, opposite angles equal, but angles 
not right angles. 

Rhomboid, having opposite sides equal, opposite angles equal, but 
angles not right angles. 

Given the Area of a Square to find its Side. 




I 


32 


ENGINEERS’ MANUAL. 


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engineers’ manual. 


33 


Rule : Extract the square root of the area. 

Formula, S^^/^rea. 

Example—Find in yards the side of a square whose surface is 
1.5 acres. Ans.—85.2 yards. 

Given the Area of a Rectangle, Rhombus, or Rhomboid, and the 
length or the perpendicular breadth to find the other dimension. 

Rule : Divide the area by the given dimension. 

A - A 

Formula, L = jj . B — —j^ 

To find any Side of a Right Angled Triangle, the other two being 
given. 

Case I. When the hypotenuse is required. 

Rule : Square the base and square the perpendicular, take the 
sum of these squares and extract the square root. 

Formula, H= \/V B 2 + P 2 ). 

Example. — The side of a square is 1200'. Find the diagonal. 
Ans. —1697 feet. 

Case II. When the perpendicular or the base is required. 

Rule : Square the hypotenuse and square the given leg, take the 
difference of these squares and extract the square root. 

The formula is deducible from the last, where 


H = ^ B 2 + P 2 
# 2 = B^+P 2 
P»- = H 2 -B 2 

.*. P = sf H 2 - B 2 
B = s/ H 2 -P 2 

Examples— 

A wall is 40' high and a ditch in front of it is 25' wide. What 
length of ladder is required to reach from the top of the wall to 
opposite side of ditch? Ans.—47.1 feet. 

At a distance of 15' a ladder 18' long is placed. How high will it 
reach ? Ans.—10' nearly. 

To Find the Area of a Regular Polygon. 

Rule I. : Multiply the length of a side by the perpendicular 
distance to the centre ; multiply the product by the number of sides 
and divide by 2, or half the perimeter multiplied by the perpendicular 
let fall from the centre to one of the sides. 

Rule II. : Square the side and multiply by the number opposite 
the name of the polygon in the last column of the following table. 

Formula, s 2 x tabular number=area. 

A polygon is a plane figure having 3 or more sides. They are 
termed regular or irregular, as the sides are equal or unequal. 






34 


ENGINEERS’ manual 


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engineers’ manual. 


35 


TABLE OF REGULAR POLYGONS. 


I. 

No. of 
of 

Sides. 

II. 

Name 

V of 

Polygon. 

III. 

Perpendicular 
Radius of 
Inscribed 
Circle. 

IV. 

Radius of 
Circumscrib¬ 
ing 

Circle 

V. 

Length of side 
Radius of Cir¬ 
cumscribed 
Circle = i. 

VI. 

Area 
Side= 1. 

3 

Eq. Triangle 

.28867 

•57735 

1.732 

.43301 

4 

Squares 

•S 

.70710 

1.414 

1.OOOO 

5 

Pentagon 

.68819 

.85065 

1•1756 

1.72047 

6 

Hexagon 

.86602 

1.0000 

I.OOOO 

2.59807 

7 

Heptagon 

1.03826 

1.15238 

.8677 


8 

Octagon 

1.20710 

1 30656 

•7653 

4 82842 

9 

Nonagon 

r -37373 

1.46190 

I684 

6 18182 

10 

Decagon 

1.53884 

1.61863 

.618 . 

7 69420 

11 

Undecagon 

1.70284 

!•77473 

•5634 - 

g.36564 

12 

Duodecagon 

1.86602 

1-93185 

•5176 . 

11.191615 


Example—What is the area of an equilateral triangle whose 
side is 30 inches ? Ans.—389.709 square inches. 

To Find the Perpendicular Height, from the centre to one of the 
sides of the Polygon, or in other words, the radius of inscribed circle, 
which is called the Apothen. 

Rule : Multiply the side by the number opposite the name in 
column III. in table. 


Example—Find the radius of an inscribed circle in the case of a 
hexagon, the side being 40. Ans. 34.6408. 

To Find the Radius of the Circumscribing Circle. 

Rule : Multiply the side by the number opposite in column IV. 
in table. 

Example—The side of an octagon is 8 inches. Find the diameter 
of the circumscribing circle. Ans.—20.90496 inches. 

To Find the Side of a Regular Polygon when the area is given. 

Rule : Divide the area by the number in last column of table 
and extract the square root. 


Side = 


y/ Area 

tabular No. 


Example—The side of a square is 3 feet, what is the side of a 
hexagon of the same area. Ans.—1.862 feet. 

To Find the Area of Irregular Polygons. 

Rule : Divide the polygon into triangles and then find the sum 
of the areas of these triangles. 

The Value of n. 

Let the radius of a circle be 1, the side of the inscribed square is 
therefore and that of the circumscribed will be equal to the dia¬ 
meter 2, hence the surface of the inscribed square will be 2 and that 
of the circumscribed will be 4. 

Let N=surface of the inscribed polygon. 

s=surface of the circumscribed polygon. 


















36 


engineers’ manual. 


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engineers’ manual. 


37 


The area of an inscribed octagon whose circumscribing circle has 
radius i = . 7653 s X4.8284271 (See table on page 35),= = 2.8284271, 

and the area of a circumscribed octagon = .8285 s X 4.8284271 = 2 + ^8 

— 3 * 3 r 37 ° 75 - 

From these inscribed and circumscribed polygons we can easily 
determine the polygons having twice the number of sides. 

.5=2.8284271 ; ^ = 3.313 7075, and the area of a polygon having 16 
sides would be = Vs + s, =3.061474 for the inscribed polygon and 
25 X 5 2 X 2.828+ X 3.3137 + 

— 5+3.061474 — 2.828 + 3.061474 — 3 - 1 825979- 

By these polygons of 16 sides the surfaces of polygons having 32 
sides can be easily determined and the process continued until the 
difference between the inscribed and the circumscribed is infinitesimal. 
Since the circle lies between these polygons it will differ from either 
polygon by less than the polygons differ from one another, and there¬ 
fore if the figures which express the areas of the two polygons agree 
they will be the true figures to express the area of the circle. 


The - following is 

the computation of these polygons carried on 

till they agree, as far as the seventh place of decimals. 

No. of Sides. 

Inscribed Polygon. Circumscribed Polygon. 

4 

2.0000000 

4.0000000 

8 

2.8284271 

3 - 3 I 37°85 

16 

3.0614674 

3.1825979 

32 

3 -I 2 I 445 I 

3 - i 5 i 7 2 49 

64 

3:1365485 

3.1441184 

128 

3 -I 4033 II 

3.1422236 

256 

3.1412772 

3.1417504 

512 

3.1415138 

3.1416321 

1024 

3 -I 4 I 5729 

3.1416025 

2048 

3 -I 4 I 5877 

3 -I 4 I 595 1 

4096 

3 -I 4 I 59 I 4 

3 -I 4 I 5933 

8192 

3 -I 4 I 5923 

3.1415928 

16384 

3 -I 4 I 5925 

3 - I 4 I 59 2 7 

32768 

3 • i 4 1 59 2 6 

3.1415926 

The approximate 

area of a circle, having 

a radius 1, is therefore 

equal to 3.1416 ; i.e. t 

, area of circle=radius 

2 X 7T=D 2 X -. It will 

A 

be observed, in the 

above table, that the 

T # 

area of the inscribed 


polygon gradually decreases as the number of sides increases, and 
the opposite with the circumscribed polygon; and it necessarily 
follows that, if the number of sides were increased infinitely, the 
two figures would ultimately agree. The above result is correct to 
seven places in decimal. 

For all practical purposes, it is generally taken as 3.1416; but, 
for very fine calculation, 3-14159265359 may be taken. 

To Find the Area of a Circle— 

Rule I.: Square the diameter, and multiply by .7854. 

Rule II.: Square the circumference, and multiply by .07958. 






38 


engineers’ manual. 




H. P. HANSEN 


T 

J. W. KREMER f 



RAILROAD, STEAMBOAT, MILL COP¬ 
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engineers’ manual. 


39 


To Find the Diameter when the Area is given— 

Rule : Divide the area by .7854 and extract the square root ; 
or, Multiply 1.12838 by the square root of the area. 

The areas of circles are to each other as the squares of their 
diameters. 

To Find the Circumference when the Area is given— 

Rule : Divide the area by .07958 and extract the square root; 

or, Circumference = v ^ rea _ 

.07958 

To Find the Area of a Circular Ring— 

Rule : ^Multiply the sum of the two diameters by their difference 
and the result by .7854. 

Formula, (D + d)(D — d) .7854 ; or, 

(D'- d *) -7834, 

when D represents the larger arid dihe smaller diameter. 

Exatnple—What is the area oT a ring formed by circles having 
their diameters 25' and 35'? Ans.—471.25 square feet. 

To Find the^Area of an Ellipse— 

Rule : Multiply the product of the two diameters by .7854. 
Formula, ( DXd ) .7854. 

Example —The two diameters of an ellipse are 30' and 25'. Find 
the area. Ans.—589.05 square feet. 

To Find the Circumference of an Ellipse or Oval— 

Rule : Multiply half the sum of the two diameters by 3.1416. 

Formula, JlfQ 3.1416 = (p + d'j 1.5708. 

Relation of the circle to its Equal, Inscribed and Circumscribed 
Square. 


Diameter of circle 
Circumference of circle 
Circumference of circle 
Diameter of circle 
Circumference of circle 
Area of circle x .9-7 
Area of circle 
Area of circle 
Side of square 
Side of square 
Perimeter of square 
Square inches 


x .88623 
X .28209 
x1.1284 
x .7071 
x .22508 
diameter 
1.2732 
.63662 
1.4142 
4.4428 
.88623 
1.2732 


== side of equal square. 

= side of equal square. 

= perimeter of equal square. 

side of inscribed square. 
f= side of inscribed square. 

= side of inscribed square. 

= area of circumscribed square. 
= area of inscribed square. 

= diarn. of circumscribed circle. 
— circum. of circumscribed circle. 
= circum. of circumscribed circle. 
= circular inches. 


To Find the Length of an Arc of a Circle— 

Rule : From 8 times the chord of half the arc subtract the chord 
of the whole arc and take one third of the remainder. 

Example—The chord of the whole arc is 18' and that of half the 
arc is 12'. What is the length of the arc ? Ans.—26 feet. 

From the height and half the chord of the arc the chord of half 





40 


engineers’ manual. 


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engineers’ manual, 


41 


the arc is determined in the same manner as finding the base or per¬ 
pendicular of a right angled triangle. 

To Find the Radius of a Circle when the chord and height of an 
Arc are given— 

Rule : Square half the chord and divide by the height, then add 
the height and divide by 2. 

Example—With what radius is a circular arch of a bridge to be 
traced whose span is 120' and rise 12.5'. Ans.—150.25 feet. 

To Find the Area of a sector of a Circle— 

Rule I.: Multiply the length of the arc by the radius and divide 
by 2. 

Example—The chord is 24/ and the height 6'. Find area of 
Sector. Ans.—208.32 square feet. 

Rule II. : Area of circle multiplied by the number of degrees in 
the arc divided by 360. 

To Find the Area of a Segment of a Circle— 

Rule : Find the area of the sector having the same arc with the 
segment, find also the area of the triangle formed by the chord of the 
segment and the two radii of the sector. If the segment be greater 
than a semicircle, take the sum of these two areas ; if the segment be 
less than a semicircle, take their difference. 


AREAS OF SEGMENTS OF A CIRCLE. 

The diameter of which is unity and supposed to be divided into 
200 equal parts. 


Height 

Area. 

Height. 

Area. 

Height. 

Area. 

Height. 

Area. 

.005 

.000471 

.130 

.059999 

•255 

.157891 

.380 

.273861 

.01 

.001329 

.135 

.063389 

.260 

.162263 

•385 

.278721 

.015 

.002438 

.140 

.066833 

• 265 

.166663 

.390 

.283593 

.02 

.003749 

.145 

.O70329 

.270 

.171090 

•395 

.288476 

.025 

.005231 

.150 

.073875 

.275 

•175542 

.400 

.293370 

.03 

.006866 

• 155 

.077470 

.280 

.180020 

.405 

.298274 

• 035 

.008638 

.160 

.081112 

.285 

.184522 

.410 

•303187 

.04 

.010538 

.165 

.084801 

.290 

.189048 

.415 

.308110 

.045 

.012555 

.170 

.088536 

•295 

•193597 

.420 

•313042 

• 05 

014681 

•175 

.092314 

.300 

.198168 

•425 

.317981 

.055 

0x69x2 

.180 

.096135 

•305 

.202762 

• 430 

.322928 

.06 

>19339 

.185 

.099997 

• 310 

.207376 

•435 

.327883 

.065 

021660 

.190 

.103910 

•315 

.212011 

.440 

.332843 

.07 

'024168 

.195 

.107843 

.320 

.216666 

•445 

.337810 

• 075 

‘026761 

.200 

.111824 

•325 

.221341 

•450 

•342783 

.08 

'029435 

.205 

.115842 

• 33 o 

. 226034 

•455 

.347760 

.085 

'032186 

.210 

.119898 

•335 

•230745 

.460 

.352742 

.09 

'035012 

.215 

.123988 

• 340 

•235473 

• 4 6 5 

.357728 

.095 

•037909 

.220 

.128114 

•345 

.240219 

•470 

.362717 

.10 

•040875 

.225 

.132273 

•350 

. 244980 

•475 

.367710 

.105 

.043908 

.230 

.136465 

•355 

.249758 

.480 

.372704 

.110 

.047006 

.235 

.140689 

• 360 

•254551 

•485 

.377701 

.115 

.050165 

.240 

•144945 

•365 

•259358 

.490 

.382700 

.120 

•053385 

.245 

.149231 

•370 

.264x79 

•495 

.387699 

.125 

.056664 

.250 

•153546 

•375 

.269OX4 

.500 

.392699 


To Calculate the Area of a Segment by the above table— 
























42 


engineers’ manual. 



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engineers’ MANUAL. 43 

Rule : Divide the height by the diameter, find the quotient in the 
column of heights, then get the corresponding area, multiply same by 
diameter of circle squared ; the product is the required area. 

Example I.—Diameter of circle, 5'; height of segment, 1'. Find 
the area of segment. 

1 -r- 5 =. 2 Area — .111824 
. 111824 x 5 2 = 2.7956 sq. ft. or 402.5664 sq. inches. 

-'s' - 



-- Example II. — Find the steam space in a boiler 6' diameter, height 
of water 4' - 3" from bottom, length 16'. 

6' -4' 3"= i' 9" ; 1' 9"6'=-^r^ — . 2916 

height .295=. 193597 
height .290=. 189048 

difference .004549 

1.6x .0045494-5 = .00151572 which added to . 189048=. 190563=Area 
of segment with a diameter of 1. 

. 190563 x 6 2 X 1 6'= 109.764288 cub. ft. 

The above shows how to calculate the area when the height of 
segment 4- diameter does not come out an exact number correspond¬ 
ing to those in table. 

To further illustrate the above The difference between the areas 
of the segments of a circle 1" in diameter . 295 and .290 high respec r 
tively is 004549, that is to say for .005 or T7r V?r difference in height, 
a difference of .004549 in area, and as our quotient in the above 

example is not .005, but .0016 the exact difference is \ of .004549 
x 1.6= .00151606, which, when added to the area of .290 which is 
. 189048=. 190563 as the correct area of circle were 1" diameter. 








44 


engineers’ manual. 


^'7Xi*X*X*X*X*X*X*X*)®SX£X5XsX*X*)@XsXsXS®®@®®®®(sX*X*X£)®(£XsX£)®®®®®0 




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engineers’ manual. 


45 

Example III.—Circle 5' diameter, segment 1.105' high. Find area. 
1.105-=-5= .22i = quotient 
height .225=. 132273 
height .220=. 128114 

difference = .004159 

.004159-1-5= .000831, which, added to . 128114=. 128945, area for 
1" diameter. 

.'. . 128945 x 5 2 =3.223625 square feet, which is absolutely cor¬ 

rect to the fourth figure in the decimals. 

We shall prove Example 1 by calculating same in a different 
manner. 

A D=c hord=4' long, A C, C D, each 2.5' long. 

Find Angle A C D — 

AC : A B :: Sin. 90 : Sin. A C B 

2 


Sin. —=.8=53° 8 ':.A C D=io6 a 16'. 
2.5 

37 



To Find the Area of a Sector when the Angle is given— 

Rule : As 360° is to the degrees in the arc of the sector, so is 
the area of the whole circle to the area of the sector ; or 
Multiply the arc of the sector by half the radius. 

First Method—360 : 106° 16' 5 2 X- 7 8 54 : Area of sector==: 

5.7956 square feet. 

From this has to be deducted the area of the triangle A C D to 
get the area of segment A D E. 

CE= 2.5'; CB—CE—B ^=2.5—1 = 1.5' 

Area of triangle A C D=2 x 1.5=3 square feet 
5.7956— 3=2. 7956=Area of sector A D E. 






Engineers’ Supply Co„ Inc 

RAILROAL AVENUE, 

Bet. Yesler Way and Washington St. 
TEL. MAIN 1191. 


DEALERS IN 


General Engineers' 
Supplies 

Lubribating Oils, Grease, 

Manhole Gaskets, Packing, 

Steam and Water Hose, 

Pipe and Fittings, and 

All kinds of Brass Goods. 


Reference: Scandinavian American Bank. 
Agents for CHAPMAN, ASHTON & BASHLIN VALVES 









engineers’ manual. 


47 

Another Example—Circle 5' diameter, segment 1.105' ,n height. 
Find area. 

I - I o 5 "i- 5 =- 22 i 
height, .225=. 132273 
height, .220=.128114 
difference=.oo4i59 

- •> i 

.004159 4-5=.000831, which, when added to .128114, gives 
. 1289455= Area for segment .221 in height, 1" diameter., 

/. . 128945 X5 9 =3.223625 square feet, which is Correct to the 
fourth place after decimal point. 

To Find the Radius of a Circle when the Chord and Height of 
the Arc are given— 

Rule : Square half the chord, and divide by the height ; then 
add the height and divide by 2. 

Example—Required the radius of a circle in which the chord is 
24', and the height of the arc is 4'. Ans.—20 feet. 

To Find the upright Surface of a Cone— 

Rule : Multiply the perimeter of the base by the slant height 
and divide by 2. 

To Find the total Surface of a Cone— 

Rule : Multiply the perimeter of the base by the slant height and 
divide by 2, then add the area of the base. 

Example I.—The radius of the base of a cone is 3'.and the slant 
height is 10'. Find the upright surface. Ans.—94.248 square feet. 

Example II.—The diameter of the base of a cone is 25' and the 
perpendicular height 40'. Find the total surface. Ans.—2136 sq. ft. 

To Find the Surface of a Sphere — 

Rule : Multiply the circumference by the diameter or square the 
diameter and multiply by 3.1416. 

Example—Find the surface of a sphere whose diameter is 12.5 
yards. Ans. —490^ square yards. 

To Find the Surface of a Cylinder— 

Rule : Multiply the length by the circumference and add the area 
of the ends. 

Example—What is the total surface of a cylinder 6' diameter and 
13.5' high ? Ans.—34.5576 square yards. 

To Find fhe Surface of a Frustum of a Cone or Pyramid— 

Rule : To the sum of the areas of both ends add the product of 
of the sum of the perimeters of the ends by one-half the slant height. 

To Find the upright surface, multiply the sum of the perimeters of 
the ends by one-half the slant height. 

Example—What is the upright surface of the frustum of a hex¬ 
agonal pyramid, the length of the sides of the ends being 20' and 12' 
'md the slant height 10'. Ans.—960 square feet. 

Find the total surface of the frustum of a cone, the circumfer- 


48 


engineers’ manual. 


THE 


acific Coast Company 


Wholesale and Retail Dealers in 


STEAM, DOMESTIC AND BLACKSMITH COALS 


MINERS OF THE CELEBRATED 

Franklin Steam Coal 


BUNKERS AND RETAIL YARD FOOT OF KING 
STREET, SEATTLE, WASH. 

TELEPHONE M IN 92. 

W. E. PEARCE, Supt. Coal Agencies. 

W. H. HAINSWORTH, Coal Agent. 








engineers’ manual 


49 

ences of whose ends are 20' and 12' and the slant height 8'. Ans.— 
171.29 square feet. 

To Find the Surface of a Wedge— 

Rule : Take the sum of all the separate surfaces. 

Similar Surfaces. 

Circles and similar plain figures are to each other as the squares 
of their diameters or of their similar sides. 

For Example—As the area of one circle is to the square of its 
diameter so is the area of another circle to the square of its diameter ; 
and as the square of the diameter of a circle is to its area so is the 
square of the diameter of another circle to its area. 

Mensuration of Solids. 

To Find the Solidity of a Sphere or Globe— 

Rule : Cube the diameter and multiply by .5236. 

Note—.5236 is one-sixth of 3.1416. 

Example—Find the solidity of a sphere 2^' in diameter. Ans.— 
8 cubic feet, 313^ cubic inches. 

To Find the Solidity of a Wedge— 

Rule : Add the three parallel edges together, multiply the sum 
by the perpendicular breadth of the base and by one-sixth of the 
perpendicular height. 

Example—The height of a wedge is if', the edge is if', length 
of base 2f', and the breadth 4^". Find the contents in cubic inches. 
Ans.—892^ cubic inches. 

To Find the Solidity of the Frustum of a Cone or Pyramid — 

Rule : Multiply the sum of the areas of the two ends and the 
mean proportional between these areas by the perpendicular height, 
and divide by 3. 

Example—What is the solidity of the frustum of a hexagonal 
pyramid, the length of the sides of the ends being 20'and 12', and 
the slant height 10'. Ans.—4888. 

To Find the Solidity of a Cube, a Parallelopipedon, a Prism or 
a Cylinder— 

Rule : Multiply the area of the base by the height. For a cube 
the rule may be put thus : Cube the sides. 

Example—A cistern is 15' long, 9' wide, and 2^' deep. How 
many gallons does it hold ? (A gallon=277.274 cubic inches.) Ans. 
— 2103^ gallons. 

A boiler 4.5' diameter is 10.5' long. How many lbs. of water will 
fill it ? (A gallon = 10 lbs.) Ans. —10407.34 lbs. 

To Find the Solidity of a Prismoid— 

Rule : To the area of the top and bottom add four times the 
area of the middle section (or the product of the sums of the length 
and breadths of the top and bottom), and multiply the sum by one- 
sixth the height. 

Note—The prismoid is a solid having parallel end areas, and 


50 


engineers’ manual. 





ASBESTOS WORK 
PUMPS 

ENGINES AND 
BOILERS 


Sweeney i Rynn 

Sanitary Plumbing , Heating 
and Ventilating Engineers 


ALL KINDS OF JOBBING A SPE¬ 
CIALTY. 


SHOP AND OFFICE, BROOKLYN 
BLDG., 

PHONE BUFF 907. 

SECOND AVE. and UNIVERSITY ST. 
SEATTLE, WASH. 





ENGINEERS MANUAL. 


S? 

may be composed of any combination of prisms, cylinders, wedges, 
pyramids or cones, or frustums of the same, whose bases and apices 
lie in the end areas. 

Example—What is the content of a trough 6' long, 3' wide at 
the top, and 5' long and 2' wide at the bottom and 2' deep? Ans.— 
27^ cubic feet. 

Similar Solids (General Principle ). 

Similar solids are to each other as the cubes of their diameters, 
sides, etc. For example—As the content of a globe is to the cube of 
its diameter, so is the content on another globe to the cube of its 
diameter ; and as the cube of the diameter of a globe is to its 
content, so is the cube of the diameter of another globe to its content. 

Cubic or Solid Measure. 

J728 cubic inches=i cubic foot. 

27 “ feet= 1 cubic yard. 

277.274 cubic inches=i gallon=io lbs. of water. 

AREAS OF SMALL CIRCLES. 

Advancing by 32nds. 


Diam. 

Area. 

Diam. 

Area. 

Diam. 

Area. 

Diam 

Area. 

X 

7 J 2 

.00076 

¥¥ 

.0621 

if 

.2216 

If 

•4793 

tV 

.0030 

tV 

.0767 

T¥ 

.2485 

if 

•5185 

¥¥ 

.0069 

ii 

.0928 

if 

.2768 

2 7 
¥¥ 

•5591 


.0122 

¥ 

. 1104 

5 

¥ 

.3068 

¥ 

.6013 

■sj 

.0192 

H 

.1296 

W 

•3382 

If 

.6450 

1 ¥ 

.0276 

T¥ 

•1503 

ii 

• 37 12 

if 

.6903 

¥¥ 

.0376 

if 

•1725 

If 

•4057 

If 

•7370 

1 

% 

.0490 

i 

.1963 

f 

• 44 x 7 

I 

•7854 


AREAS OF CIRCLES, 
Advancing by 8ths. 


AREAS. 


Diam. 11 

0 

A 

X 

X 

X 

X 

X 

X 

a 

rS 

S 

O 

.0 

.0122 

.0490 

. 1104 

- i 9 6 3 

.3068 

• 44 x 7 

• 6013 

0 

I 

•7854 

.9940 

.1227 

1.484 

1.767 

2.073 

2.405 

2.761 

1 

2 

3 1 4 l6 

3-546 

3-976 

4-430 

4 908 

5 - 4 11 

5-939 

6.491 

2 

3 

7.068 

7.669 

8.295 

8.946 

9.621 

10.32 

11.04 

11.79 

3 

4 

12.56 

I 3 - 3 6 

14.18 

i 5- 0 3 

i 5 - 9 o 

16.80 

17.72 

18 66 

4 

5 

19.63 

20.62 

21.64 

22.69 

23-75 

24.85 

25.96 

27.10 

5 












































52 


ENGINEERS’ manual. 




lYlita! 

of 



SAN FRANCISCO, CAL. and SEATTLE, WN. 
Manufacturers of All Grades of 


Anti-Attrition Metals 

SAN FRANCISCO, CAL., 313 and 315 Howard St. 
SEATTLE, WASH., Cor. 1st Ave. So. and Atlantic St 


Armature Lining Metal has no equal for Armatures, En¬ 
gine Bearings, Planers and any heavy work where there is a 
pound and vibration. Some of the electric plants using Arma¬ 
ture: 

Market St. System, San Francisco, Cal. 

Los Angeles Ry. Co., Los Angeles, c*l. 

Seattle Electric Co., Seattle, Wash. 

Vancouver Electric Co., Vancouver, B. C 

Victoria Electric Co., Victoria, B. C. 

Some of the Saw Mills: 

St. Paul & Tacoma Lumber Co. 

Tacoma Mill Co. 

Puget Mill Co., Etc. 

Correspondence Solicited. 




engineers’ manual. 


53 


Areas of Circles.— Continued. 


1 

•fH 

O 

y& 

X 


X 

H 

X 

ft 

s' 

A 

6 

28. 

27 

29 

46 

3° 

67 

3 1 -9 1 

33 -i 8 

34 

47 

35-78 

37- 12 

6 

7 

38.48 

39 

87 

4 1 • 

28 

42.71 

44.17 

45 

66 

47.17 

48 70 

7 

8 

5° 

26 

5 1 • 

84 

53- 

45 

55-08 

56-74 

58 

42 

60.13 

61.86 

8 

9 

63 

6l 

65- 

39 

67 

20 

69.02 

70 88 

72 

75 

74.66 

76.58 

9 

10 

78. 

54 

80 

5i 

82. 

51 

84 54 

86.59 

88 

66 

90.76 

92.88 

IO 

11 

95 

03 

97 

20 

99 

40 

101 6 

103.8 

106 

I 

108.4 

no.7 

I I 

12 

H3- 

O 

”5- 

4 

117. 

8 

102 2 

122.7 

125 

I 

127.6 

130.1 

12 

13 

132. 

7 

i35- 

2 

137- 

8 

140 5 

143-i 

H5 

8 

148.4 

I5 1 -2 

13 

14 

i53- 

9 

156. 

6 

159- 

4 

162 2 

165.1 

167 

9 

170.8 

*73 7 

14 

*5 

176 

7 

179 

6 

182 

6 

185.6 

188.6 

I 9 I 

7 

194.8 

T 97 -9 

15 

l6 

201 

O 

204 

2 

207 

3 

210 5 

213.8 

217 

O 

220.3 

223.6 

l6 

J 7 

226 

9 

230 

3 

233 

7 

237.1 

240 5 

243 

9 

247.4 

250.9 

r 7 

18 

254 

4 

258 

O 

261 

5 

265.1 

268 8 

272 

4 

276 1 

279.8 

18 

*9 

283 

5 

287 

2 

291 

O 

294.8 

298.6 

302 

4 

3°6-3 

3 IO -2 

19 

20 

3 H 

I 

318 

I 

322 

O 

326 0 

33° 

334 

I 

338-1 

342.2 

20 

21 

34 6 

3 

350 

4 

354 

6 

358 8 

363-0 

367 

2 

37 1 -5 

375 8 

21 

22 

380 

I 

384 

4 

388 

8 

393 2 

397 6 

402 

O 

406.4 

410 9 

22 

23 

4 T 5 

4 

420 

O 

424 

•5 

429.1 

433-7 

438 

3 

443 0 

447.6 

23 

24 

452 

3 

457 

I 

461 

.8 

466.6 

47 1 - 4 

476 

2 

481.1 

485 9 

24 

25 

490 

8 

495 

7 

5°° 

7 

5°5-7 

5 IO -7 

5 T 5 

7 

520.7 

525-8 

25 

26 

530 

9 

536 

O 

54i 

I 

546 3 

55 1 - 5 

556 

7 

562.0 

567.2 

29 

27 

572 

5 

577 

8 

583 

2 

588 5 

593-9 

599 

3 

604 8 

610.2 

27 

28 

615 

7 

621 

.2 

626 

7 

632-3 

637 9 

643 

5 

649.1 

654.8 

28 

29 

660 

5 

666 

2 

671 

•9 

677-7 

683 4 

689 

2 

695.1 

700 9 

29 

30 

706 

8 

7 12 

7 

718 

6 

724.6 

730 6 

736 

6 

742.6 

748.6 

30 

3 1 

754 

8 

760 

9 

767 

O 

773 1 

779-3 

785 

5 

791.7 

798.0 

31 

32 

804 

2 

810 

■5 

816 

9 

823 2 

829.6 

836 

0 

842.4 

848.8 

32 

33 

855 

• 3 

861 

.8 

868 

3 

874.8 

881.4 

888 

0 

894.6 

9 OI -3 

33 

34 

907 

■9 

914 

6 

921 

3 

928 1 

934-8 

94 1 

6 

948-4 

955-3 

34 

35 

962 

I 

969 

.O 

975 

•9 

982.8 

989.8 

996 

8 

1003 8 

1010 8 

35 

36 

1017 

9 

1025 

O 

1032 

I 

1039.2 

1046.4 

1053 

■5 

1060.7 

1068.0 

36 

37 

io 75 

2 

1082 

•5 

1089 

.8 

1097.1 

1104 5 

I I I I 

.8 

I I 19 2 

1126.7 

37 

38 

ii34 

I 

I I4I 

.6 

1149 

I 

1156 6 

1164 2 

1171 

•7 

1179-3 

1186.9 

38 

39 

1194 

6 

1202 

•3 

1210 

.O 

1217 7 

1225 4 

1233 

2 

I24I.O 

1248.8 

39 

40 

1256 

6 

1264 

5 

1272 

•4 

1280 3 

1288 3 

1296 

2 

1304 2 

x 3 12 2 

40 

4i 

1320 

3 

1328 

3 

133 6 

4 

1344-5 

I352-7 

1360 

.8 

1369.0 

1377.2 

4 1 

42 

1385 

4 

1393 

7 

1402 

O 

i4 IO -3 

1418 6 

1427 

.O 

H35 4 

1443.8 

42 

43 

i45 2 

2 

1460 

•7 

1469 

. I 

1477.6 

1486.2 

1494 

•7 

1503-3 

15 11 9 

43 

44 

1520 

5 

T 5 2 9 

2 

1537 

•9 

1546 6 

1555-3 

1564 

O 

1572.8 

1581.6 

44 

45 

!59° 

4 

1599 

3 

1608 

. 2 

1617.0 

1626.0 

1634 

•9 

1643.9 

1652 9 

45 

46 

1661 

■9 

1671 

.0 

1680 

.O 

1689.1 

1698.2 

1707 

4 

1716 5 

r 7 2 5-7 

46 

47 

1734 

9 

1744 

. 2 

x 753 

•5 

1762.7 

1772. 1 

1781 

•4 

1790.8 

1800.1 

47 

48 

1809 

6 

1819 

.0 

1828 

5 

1837-9 

1847 5 

1857 

.O 

1866 6 

1876.1 

48 

49 

1885 

7 

1895 

4 

1905 

O 

I 9 I 4-7 

1924-4 

1934 

. 2 

1943 9 

I 953 7 

49 

5° 

1963 

5 

1973 

3 

1983 

. 2 

1993 - 1 

2003 O 

2012 

9 

2022.8 

2032.8 

5° 

5i 

2042 

8 

2052 

•9 

2062 

•9 

2073.0 

2083 1 

2093 

. 2 

2103.4 

12113-5 

5i 















































ENGINEERS’ manual. 


Issaquah Coal Co. 

OUR WASHED SCREENINGS MAKE MORE 
STEAM WITH LESS LABOR, AND AT 
LESS COST THAN ANY FUEL 
ON THE MARKET. 


OUR LUMP CAN BE ADVANTAGEOUSLY 
USED WHERE IT IS NECESSARY 
TO FORCE THE BOILERS. 


ISSAQUAH COAL CO. 

SEATTLE, WASH. 

PHONE MAIN 976 


MINES AT ISSAQUAH, WASH. 
CAPACITY 750 TONS DAILY 








engineers’ manual. 


55 


Areas of Circles .— Continued. 


1 Diam. 11 

0 

X 

X 

X 

X 

X 

X 

X 

S 

Sg 

5 

52 

2123.7 

21339 

2144.2 

2154-5 

2164.8 

2175.I 

2185.4 

2195 

.8 

52 

53 

2206.2 

2216.6 

2227.1 

2237-5 

2248 0 

2258.5 

2269.I 

2279 

.6 

53 

54 

2290.2 

2300.8 

2311-5 

2322.1 

2332.8 

2343 5 

2354-3 

2365 

.0 

54 

55 

2375-8 

2386.6 

2397-5 

2408.3 

2419.2 

2430.2 

2441.I 

2452 

.0 

55 

56 

2463.0 

2474.0 

2485.1 

2496 1 

2507•2 

2518 3 

2529.4 

2540 

6- 

S 6 

57 

2551-8 

25630 

2574 2 

2585-5 

2596.7 

2608.0 

2619.4 

2630. 

•7 

57 

5« 

2642.1 

2653-5 

2664.9 

2676.4 

2687.8 

2699.3 

2710.9 

2722. 

■4 

58 

59 

2734.0 

2745.6 

2757.2 

2768.8 

2780.5 

2792.2 

2803.9 

2815. 

■7 

59 

60 

2827.4 

2839.2 

2851.1 

2862.9 

2874.8 

2886.7 

2898 6 

2910. 

•5 

60 

61 

2922.5 

2934-5 

2946.5 

2958-5 

2970.6 

2982.7 

2994.8 

3006. 

9 

61 

62 

3 OI 9- 1 

3031-3 

3043-5 

3055-7 

3068.0 

3080.3 

3092.6 

3 io 4 

•9 

62 

63 

3ii7-3 

3129 6 

3142.0 

3I54-5 

3 i6 6 9 

3 J 79 4 

3 T 9 J -9 

3204 

4 

63 

64 

3217.0 

3229.6 

3242.2 

3254-8 

3267-5 

3280.1 

3292 8 

3305- 

6 

64 

65 

3318.3 

333i-i 

3343-9 

3356.7 

3369.6 

3382 4 

3395-3 

3408 

3 

6s 

66 

3421.2 

3434-2 

3447-2 

3460.2 

3473-2 

34863 

3499-4 

35 12 . 

5 

66 

67 

3525-7 

3538.8 

3552 0 

3565-2 

3578-5 

359 1 -7 

3605.0 

3618. 

4 

67 

68 

3631-7 

3645 • 1 

36584 

3671.9 

36853 

3698.8 

3712.2 

3725- 

8 

68 

69 

3739-3 

3752-8 

3766.4 

3780.9 

3793•7 

3807.3 

3821.0 1 

3834- 

7 

69 

70 

3848.5 

3862.2 

3876.0 

3889.8 

3903.6 

39I7-5 

3931-4 

3945- 

3 

7o 

7i 

3959-2 

3973-2 

3987-I 

4001.1 

4015.2 

4029.2 

4043•3 

4057- 

4 

7 1 

72 

4°7 r -5 

4085.7 

4099.8 

4114.0 

4128.3 

4 i 4 2. 5 

4 i 5 6 -8 

4171. 

1 

72 

73 

4185-5 

4199.7 

4214.1 

4228.5 

4242.9 

4257-4 

4271.8 

4286. 

3 

73 

74 

4300.9 

43I5-4 

4330.0 

4344-6 

4359-2 

4373-8 

4388.5 

4403- 

,2 

74 

75 

44 r 7-9 

4432.6 

4447.4 

4462.2 

4477.0 

4491.8 

4506.7 

4521. 

6 

75 

76 

4536.5 

455 1 -4 

4566.4 

458 i -3 

4596-4 

4611.4 

4626.4 

4641. 

5 

76 

77 

4656.6 

4671.8 

4686.9 

4702.1 

47I7-3 

4732-5 

4747.8 

4763- 

1 

77 

78 

4778.4 

4793-7 

4809.1 

4824.4 

4839.8 

4855-3' 

4870.7 

4886. 

2 

78 

79 

4901.7 

49 I 7- 2 

4932.8 

4948.3 

4963.9 

4979-5 

4995.2 

5 OI °- 

9 

79 

80 

5026.6 

5 0 42-3 

5058.0 

5073-8 

5089.6 

5 io 5-4 

5121.2 

5 r 37- 

1 

80 


CIRCUMFERENCES OF CIRCLES, 
Advancing- by 8ths. 


Circumferences. 


i 

p 

0 

X 

X 

X 

X 

X 

X 

X 

a 

5 

0 

.0 

• 3927 

•7854 

1.178 

1-57° 

1.963 

2-356 

2.740 

0 

1 

3- I 4 I 

3-534 

3-927 

4-3I9 

4.712 

5-105 

5-497 

5.890 

1 

2 

6.283 

6.675 

7.068 

7.461 

7-854 

8.246 

8.639 

9.032 

2 

3 

9.424 

9.817 

10.21 

10.60 

10.99 

11.38 

11.78 

12.17 

3 

4 

12.56 

12 -95 

13-35 

13-74 

i 4- x 3 

! 4 - 5 2 

14.92 

15-31 

4 

5 

i 5 - 7 o 

16.10 1 

16.49 

16.88 

17.27 

J 7 - 67 

18.06 

18.45 

5 

















































56 


engineers’ manual. 


HOISTING ENGINES 

... OF ... 

ALL KINDS A SPECIALTY. 

FOR LOGGING, MINING, PILEDRIVING, CON¬ 
TRACTORS’ USE, RAILROAD PUR¬ 
POSES, ETC. 

All our engines are built to gauges and templates 
on the duplicate part system. They are best in de¬ 
sign, most durable and economical. 

WE INVITE CORRESPONDENCE. 


W ashington 
Iron Works Co. 


SEATTLE, WASH. 


engineers’ manual. 


57 


Circumferences.- Continued . 


Diam. || 

0 

# • 

. > - 


X 

X 

'H 

V * 

S 

3 

6 

18.84 

19.24 

19.63 

20.02 

20.42 

20.81 

21.20 

21.59 

6 

7 

21.99 

22.38 

22.77 

23.16 

23-56 

23-95 

24.34 

24.74 

7 

8 

2513 

25-52 

25.91 

26.31 

26.70 

27.09 

27.48 

27.88 

8 

9 

28.27 

28.66 

29.05 

29-45 

29.84 

30-23 

30-63 

3 ‘ 02 

9 

IO 

3 ‘- 4 ‘ 

3 1 -8o 

32.20 

3259 

32.98 

33-37 

33-77 

34.16 

10 

11 

34-55 

34-95 

35-34 

35-73 

36.12 

36.52 

36.91 

37-30 

11 

12 

37 69 

38.09 

38.48 

38.87 

39-27 

39.66 

40.05 

40.44 

12 

13 

40.84 

4 1 • 23 

41.62 

42.01 

42.41 

42.80 

43-19 

43.58 

‘3 

14 

43 - 9 8 

44-37 

44.76 

45 - J 6 

45-55 

45-94 

46.33 

46.73 

‘4 

i 5 

47.12 

47 - 5 1 

47.90 

48.30 

48.69 

49.08 

49.48 

49.87 

‘5 

16 

50.26 

50-65 

5 1 -°5 

5 1 -44 

51-83 

52.22 

52.62 

53-01 

16 

17 

53-40 

53-79 

54-‘9 

54-58 

54-97 

55-37 

55-76 

56-15 

‘7 

18 

56-54 

59-94 

57-33 

57-72 

58.11 

58.51 

58.90 

59-29 

18 

‘9 

59- 6 9 

60.08 

60.47 

60.86 

61.26 

61.65 

62.04 

62.43 

‘9 

20 

62.83 

63.22 

63.61 

64.01 

64.40 

64.79 

65.18 

65-58 

20 

21 

65 97 

66.36 

66.75 

67-15 

67-54 

67 93 

68.32 

68.72 

21 

22 

69.11 

69.50 

69.90 

70.29 

70.68 

7 1 -°7 

71.47 

71.86 

22 

2 3 

7 2 - 2 5 

72.64 

73 04 

73-43 

73-82 

74.22 

74.61 

75.00 

23 

24 

75-39 

75-79 

76.18 

76.57 

76.96 

77-36 

77-75 

78.14 

24 

25 

7854 

78-93 

79-32 

79 - 7 1 

80.10 

80.50 

80.89 

81.28 

25 

26 

81.68 

82.07 

82.46 

82.85 

83-25 

83.64 

84.03 

84-43 

26 

27 

84.82 

85-21 

85.60 

86.00 

86.39 

86.78 

87.17 

87-57 

27 

28 

87.96 

88.35 

88.75 

89.14 

89-53 

89.92 

9032 

90.71 

28 

29 

91.10 

9 1 -49 

91.89 

92.28 

92.67 

93.06 

93.46 

93-85 

29 

30 

94.24 

94.64 

95-°3 

95-42 

95.81 

96.21 

96.60 

96.99 

30 

3 T 

97-4 

97.8 

98.2 

98.6 

99.0 

99.4 

99-7 

100.1 

3 1 

32 

100.5 

100.9 

101.3 

101.7 

102.1 

102.5 

102.9 

1033 

32 

33 

103.7 

104.1 

104.5 

104.9 

105.2 

105.6 

106.0 

106.4 

33 

34 

106.8 

107.2 

107.6 

108.0 

108.4 

108.8 

109.2 

109.6 

34 

35 

110.0 

no.3 

no.7 

n 1.1 

111 -5 

1 n .9 

112.3 

112.7 

35 

3 6 

113.1 

.11 3’-5 

“ 3-9 

H 4-3 

114.7 

“ 5 - 1 

“ 5-5 

115.8 

36 

37 

116.2 

116.6 

117.0 

H 7-4 

117.8 

118.2 

118.6 

119.0 

37 

38 

119.4 

119.8 

120.2 

120.6 

121.0 

121.3 

121.7 

122.1 

38 

39 

122.5 

122.9 

123.3 

• 23-7 

124.1 

124.5 

124.9 

‘ 25-3 

39 

40 

125.7 

126.1 

126.4 

126.8 

127.2 

127.6 

128.0 

128.4 

40 

4 1 

128.8 

129.2 

129.6 

130.0 

130.4 

130.8 

131-2 

‘ 3 ‘ - 6 

4 i 

42 

13 1 -9 

132-3 

.32.7 

‘ 33 - 1 

‘ 33-5 

‘ 33-9 

‘ 34-3 

‘ 34-7 

42 

43 

135 - 1 

135-5 

135-9 

136-3 

136.7 

‘ 37-1 

137-4 

‘ 37-8 

43 

44 

138.2 

138.6 

139.0 

139-4 

139.8 

140.2 

140.6 

141.0 

44 

45 

141.4 

141.8 

142.2 

142.6 

142.9 

‘ 43-3 

‘ 43-7 

144.1 

45 

46 

J 44-5 

144.9 

H 5-3 

i 45 -7 

146.1 

146.5 

146.9 

147-3 

46 

47 

147.7 

148.0 

T48.4 

148.8 

149.2 

149.6 

150.0 

150.4 

47 

48 

150.8 

151.2 

151.6 

152.0 

152.4 

152.8 

‘ 53-2 

153-5 

48 

49 

1539 

154-3 

154-7 

‘55 1 

‘ 55-5 

‘ 55-9 

1563 

156-7 

49 

50 

5 1 

157 1 
160.2 

157-5 

160.6 

157-9 
161.0 

158-3 

161.4 

1587 

161.8 

159.0 

162.2 

‘ 59-4 

162.6 

159.8 

163.0 

5 ° 

51 



























58 


engineers’ manual. 


ME BRIDGE (0. 



Telephone Main 509. 
PACIFIC BLOCK, SEATTLE, 


HYDRAULIC DREDGING, BRIDGES, 
WHARVES, HARBOR IMPROVEMENTS 
ALL KINDS OF HEAVY CONSTRUCTION 














engineers’ manual. 


59 


Circumferences.— Continued . 


A 

A 

c 3 

s 

0 

% 

X 

Vs 

V 

H 

u 


s 

c6 

5 

5 2 

163.4 

163.8 

164.1 

l6 4-5 

164.9 

165.3 

165.7 

166.1 

52 

53 

166.5 

166.9 

167.3 

167.7 

168.1 

168.5 

168.9 

169.3 

53 

54 

169.6 

170.0 

170.4 

170.8 

17 I . 2 

171.6 

172.0 

172.4 

54 

55 

172.8 

173.2 

173.6 

174.0 

J 74-4 

174.8 

175-1 

175-5 

55 

5 6 

' 75-9 

176.3 

176.7 

177.1 

177-5 

T 77-9 

178.3 

178.7 

56 

57 

179. 1 

179-5 

179.9 

180.2 

180.6 

181.0 

181.4 

181.8 

57 

5 8 

182,. 2 

182.6 

183.0 

183.4 

183.8 

184.2 

184.6 

185.0 

58 

59 

185.4 

■85-7 

186.1 

186.5 

186.9 

187.3 

187.7 

188.1 

59 

60 

188.5 

188.9 

189.3 

189.7 

190.1 

190.5 

I 9°-9 

191.2 

60 

61 

191.6 

192.0 

192.4 

192.8 

193.2 

193.6 

194.0 

194.4 

61 

62 

194.8 

195.2 

195.6 

196.0 

196.4 

196.7 

T 97 - 1 

197-5 

62 

63 

197.9 

198.3 

198.7 

199.1 

, 199-5 

199.9 

200.3 

200.7 

6 3 

64 

201.1 

201.5 

201.8 

202.2 

202.6 

203.0 

203.4 

203.8 

64 

65 

204.2 

204.6 

205.0 

205.4 

205.8 

206.2 

206.6 

207.0 

65 

66 

207.3 

207.7 

208.1 

208.5 

On 

06 

O 

N 

209.3 

209.7 

216.1 

66 

67 

210.5 

210.9 

211.3 

211.7 

212. I 

212.5 

212.8 

213.2 

67 

68 

213.6 

214.0 

214.4 

214.8 

215.2 

215.6 

216.0 

216.4 

68 

69 

216.8 

217.2 

217.6 

217.9 

2l8.3 

218.7 

219.1 

219.5 

69 

70 

219.9 

220.3 

220.7 

221.1 

221.5 

221.9 

222.3 

222.7 

70 

7 1 

223.1 

223.4 

223.8 

224.2 

224.6 

225.0 

225.4 

225.8 

7 * 

7 2 

226.2 

226.6 

227.0 

227.4 

227.8 

228.2 

228.6 

228.9 

7 2 

73 

229.3 

229.7 

230.1 

230-5 

2309 

231-3 

231.7 

2321 

73 

74 

232-5 

232.9 

233-3 

233-7 

234.O 

234-4 

234.8 

235-2 

74 

75 

235 -6 

236.0 

236.4 

236.8 

237.2 

237.6 

238.0 

238.4 

75 

76 

238.8 

239.2 

239-5 

239-9 

24O.3 

240.7 

241.1 

241.5 

76 

77 

24 1 9 

242.3 

242.7 

243.1 

243-5 

243-9 

244-3 

244.7 

77 

78 

245.0 

245-4 

245.8 

246.2 

246.6 

247.0 

247.4 

247.8 

78 

79 

248.2 

248.6 

249.0 

249.4 

249.8 

250.1 

2505 

250.9 

79 

80 

25 i -3 

251-7 

252.0 

252-5 

252.9 

253-3 

253-7 

254 • 1 

80 


FOAMING IN BOILERS. 

The causes are (i) dirty water, (2) trying to evaporate more 
water than the size and construction of the boiler is intended for, 
(3) taking steam too low down, (4) insufficient steam room, (5) imper¬ 
fect construction of boiler, (6) too small a steam pipe, (7) and some¬ 
times by carrying the water line too high. 

Too little attention is paid to boilers with regard to their evapo¬ 
rative power. Where the boiler is large enough for the water to 
circulate, and there is enough surface to give steam, foaming 
never occurs. 

As the particles of the steam have to escape to the surface of the 
water in the boiler, unless that is in proportion to the amount of steam 
to be generated, it will be delivered with such violence that the water 
will be mixed with it, and cause foaming. 


























60 


engineers’ manual. 



Fai rban ks=florse 

Gasoline Engines 


FROM 1 y 2 TO 150 HORSE-POWER. 

SIMPLE, DURABLE, RELIABLE 

CALL AND EXAMINE THEM. 


ALSO STEAM PUMPS, 

STEAM ENGINES, 

BOILERS, ETC. 


GEO. B.ADAIR&SON 

AGENTS. 

309 Occidental Ave., SEATTLE. 


-oo>zo-)>—ICO 








ENGINEERS’ MANUAL. 6! 

For violent ebullition, a plate hung over the hole when the steam enters 
the dome from the boiler is a good thing and prevents a rush of water by 
breaking it, when the throttle is opened suddenly. 

In cases of very violent foaming it is imperative to check the draft and 
cover'fires. 

The steam pipe may be carried through the flange six inches into the 
dome, which will prevent the water entering the pipes by following the sides 
of the dome as it does. 

A case of priming was stopped by removing some of the tubes under the 
smokestack in the U. S. Steamer Galina, and substituting bolts. 

Clean water, plenty of surface, plenty of steam room, large steam pipes, 
boilers large enough to generate steam without forcing the fires, are all that is 
required to prevent foaming. 

Qas and Oil Engines. 

Before starting up a gac or oil engine it is always well to look 
it over thoroughly in order to make sure that all the adjustments 
are properly made. First, test the igniter by turning the engine 
over very slowly until the snap of the spark in the cylinder is 
heard, and note the position of the crank when this occurs; it 
should be very nearly on the inner center. In making this test 
it will be found convenient to prop open the exhaust valve and 
open a pet cock through which to listen for the snap of the spark. 
If it be found difficult to detect the exact instant when the 
igniter flashes then the tripping of the firing cam on the outside 
must be taken as the guide. In this case, the position of the 
crank after the igniter has tripped is the correct one to note, 
and not the position when the cam begins to move the trigger. 
This applies, obviously, to electric ignition only unless the 
engine is equipped with a timing valve in connection with an 
ignition-tube, which is seldom the casein this country. 

Next inspect all of the valves, making sure that they open and 
close at the right parts of the stroke. The catalogue or card of 
instructions sent out by the maker will state just what should be 
expected in this particular. The mechanism of the engine being 
all right, oil it thoroughly and if a hot tube igniter is used, heat 
it to a bright red—not white—and turn the engine over twice 
with the gas valve about half way open (“open” meaning the 
position of running); the valve is usually provided with a dial 
on which the starting position is marked. If there is no dial, 
and the engine is starting for the first time, one man will have to 
turn the fly wheel over while another opens the gas valve a little 
at a time until an explosion is secured. As soon as the first ex¬ 
plosion is obtained, open the gas valve slowly, but steadily, as 
the engine gains speed until it ceases to increase its rate of rota¬ 
tion. A little care will have to be observed to avoid “choking” 
the charge, i. e ., giving so much gas as to prevent ignition. 
Should the engine slacken in speed after once beginning to ac¬ 
celerate, it is an indication that too much gas is being admitted 
for the quantity of air, and the gas valve should be closed until 
acceleration is resumed. 

Immediately upon the attainment of normal speed by the 
engine the jacket water should be turned on—not before. Reg¬ 
ulate the flow until the issuing jacket water is about blood warm 



62 


engineers’ manual. 


De Solla-Deussing Co. 

315-317 First Avenue So., Seattle, Wash. 

ASBESTOS GOODS 



Boiler and Pipe Covering 



There is only one covering, and it bears the 
label: “ Keasby & Mattison’s 85 % Magnesia. 









63 


engineers’ manual 

and then throw on the load. If the load is a fairly constant one, 
the jacket water flow can be then adjusted until the issuing 
water is just as hot as one’s hand can bear; it is much bettei, 
however, to use a thermometer and adjust the flow until the k - 
suing water is about i8o° Fah. (This is hotter than one can 
stand with the naked hand, of course, and it is, therefore, better 
Louse the thermometer as the guide, because the higher the 
temperature of the issuing jacket water, up to certain limits, tie 
better the efficiency of the engine.) The thermometer should 
not be put directly in the water, but set in a cup of oil, which in 
turn sets in a tee in the delivery pipe. The cup can easily be 
made to screw into the side of the tee and be left there perma¬ 
nently; in this case the tee should be a couple of sizes larg- r 
than the pipe, so that the oil cup will not throttle the flow of the 
jacket water too greatly. 

If the load is a fluctuating one, then the jacket water flow 
should be adjusted so that at maximum load the temperature of 
the issuing water is about 200° Fah. 

If hot tube ignition be employed, the burner should be regu¬ 
lated so as to maintain the tube at a bright cherry red and no 
hotter. Any higher temperature simply shortens the life of the 
tube and does not increase the reliability of the ignition a par 
tide. If electric ignition be used, a good reliable form of closed- 
circuit battery should be installed. The ordinary dry batteries 
and sal ammoniac cells used for electric bells and telephones are 
entirely unsuited to gas engine service for the reason that they 
become exhausted (polarized) too soon under continuous opera¬ 
tion. A greater number of closed-circuit cells will be required, 
but the results will justify it abundantly. 

Any good machine oil will do for the bearings of the engine, 
but the cylinder should never be abused by serving to it anything 
but the special oil made for that purpose. ' The very best grade 
of cylinder oil used in steam engines is about the worst that can 
be used in the cylinder of a gas engine for the reason that it clots 
and carbonizes under the intense dry heat. If the engine is of 
the two-part cycle type with a closed crank case, then crank- 
ca-e oil and no other should be used for that part of the engine. 

All gas or gasoline engines should be provided with a cock in the gas inlet 
pipe, having a graduated dial, by means of which the delivery of gas may be 
adjusted and noted independently of the air supply, excepting those engines 
in which the governor controls the admission of gas but not that of air. Bar¬ 
ring these engines, the attendant should always adjust his gas cock so that at 
normal load the governor weights are thrown out to their maximum distance 
from the center of rotation. The precaution must be taken, however, not to 
throttle the gas supply to such an extent that the charge will not explode. In 
the case of a hit-and-miss engine, it will be found good practice to adjust the 
gas supply so that at no load the engine will take a charge once in about 
every four complete cycles, or to throttle the supply just as far as possible 
without causing failure to ignite. It may be necessary to open the valve a 
little at full load, but it probably will not be. This can be determined, of 
course, by noting the speed of the engine; if it drops under full load, then the 
valve must be opened a trifle at a time until the engine will maintain its speed.' 

In shutting down, the best order of procedure is to (1) shut oflf the gas sup¬ 
ply; (2) disconnect the igniter, if electric, or turn out the burner, if hot-tube; 
(3) shut oflf the jacket water, and (4) shut oflf the oil feed. 







64 


ENGINEERS’ manual. 


Ilensbav. Dulhley 
Ho. 

ENGINEERS 

I MACHINERY 

AND SUPPLIES 

IRON AND WOODWORKING TOOLS. 

‘i* 

315 and 317 FIRST AVE. SOUTH, 


TELEPHONE MAIN 573. SEATTLE. 









engineers’ manual 65 

USEFUL INFORMATION. 

The height of a column of fresh water, equal to a pressure of 1 
lb. per square inch, is 2.31 feet. A column of water 1 foot high exerts 
a pressure of .433 lbs. per square inch. The capacity of a cylinder 
in gallons is equal to the length in inches multiplied by the area in 
inches, divided by the cubical contents of one gallon in inches (see 
following table). The velocity in feet per minute, necessary to dis¬ 
charge a given volume of water in a given time, is found by multiply¬ 
ing the number of cubic feet of water by 144 and dividing the product 
by the area of the pipe in inches. The area of a required pipe, the 
volume and velocity being given, is found by multiplying the number 
of cubic feet of water by 144 and dividing the product by the velocity 
in feet per minute. The area being found, the diameter is obtained 
by the Table of Areas. Doubling the diameter of the pipe increases 
its capacity four times. The friction of liquids in pipes increases as 
the square of the velocity. The horse-power necessary to elevate 
water to a given height is found by multiplying the weight of the 
water elevated per minute, in pounds, by the height in feet and divid¬ 
ing the product by 33,000. An allowance of 25 per cent, should be 
made for friction, etc. 


Weight and Capacity of Different Standard Gallons of 

Water. 



Cubic 
Inches in a 
Gallon. 

Weight of 
a Gallon 
in lbs. 

Gallons in 
a cubic 
foot. 

Weight of a cu¬ 
bic foot of water, 
English stand¬ 
ard, 62.321 pounds 
Avoirdupois. 

Imperial or English. 

United States. 

277.274 

231 

10.00 

8.33m 

6.232102 

7.480519 



A cubic inch of water, evaporated under ordinary atmospheric 
pressure will be converted into approximately 1 cubic foot of 
steam, and it exerts a mechanical force equal to lifting 2,120 lbs. 
1' high. 27,222 cubic feet of steam weigh 1 lb, 13,817 cubic feet of 
air weigh 1 lb., the specific gravity of steam, at atmospheric 
pressure being .441 that of air at 34 0 F., and .0006 that of water at 
the same temperature. 

The government method prescribed for cleaning brass, and in use 
at all the United States arsenals, is said to be the best in the world. 
The plan is to make a mixture of two parts nitric and one part sul¬ 
phuric acid in a stone jar, having also a pail of fresh water and a box 
of sawdust. The articles to be treated are first dipped into the acid, 
then placed in the water, and finally rubbed with the sawdust This 
immediately changes them to a brilliant color. If the brass is greasy, 
it is first dipped into a strong solution of potash or soda in warm 
water, and then rinsed. This dissolves the grease and leaves the 
acid free to act. 

In backing out bolts, without protection for the thread, strike the 
hardest blows possible with a heavy hammer. Light blows with a 
light hammer only upset the bolt. 




















66 


engineers’ manual. 


,77. C Zouseyti Co. 

telephone Wain 1180 

LEATHER AND FINDINGS 


Maltese Cross Rubber Belting, 

Red Strip Rubber Belting, 

Pyramid Brand Leather Belting, 
Dorman Patent Wood Split Pulleys, 
Plomo Belt Dressing, 

Sheet Packings, 

Piston Packings, 

Gaskets of all kinds. 
Water, Steam and 

Suction Hose. 


Lace Leather, Emery Wheels 


316 SECOND AVE. SOUTH, SEATTLE, WASH. 
1511 PACIFIC AVE., TACOMA, WASH. 








engineers’ manual. 


67 


BELTING. 

Although there is not nearly as much known in general about 
the power of transmitting agencies as there should be. still it seems 
that almost any other method or means is better understood than 
belts. One of the chief difficulties in the way of a better knowledge 
of the belting problem, is the relations that belts and pulleys bear to 
each other. The general supposition, and one that leads to many 
errors, is that the larger in diameter a pully is, the greater its holding 
capacity—the belt will not slip so easily, is the belief. But it is merely 
a belief, and has nothing to sustain it unless it be faith, and faith 
without work is an uncertain factor. I would like here to impress 
upon the minds of all interested the following immutable principles 
or laws : 

1. The adhesion of the belt to the pulley is the same, the arc or 
> number of degrees of contact, aggregate tension or weight being the 

same , without reference to width of belt or diameter of pulley. 

2. A belt will slip just as readily on a pulley f in diameter as it 
will on a pulley 2' in diameter, provided the conditions of the faces of 
the pulleys , the arc of contact, the tension and the number of feet the 
belts travel per minute are the same in both cases. 

3. A belt of a given width, and making 2,000 or any other given 
number of feet per minute, will transmit as much power running on 
pulleys, f in diameter, provided the arc of contact, tension and con¬ 
ditions of pulley faces all be the same in both cases. 

It must be remembered, in reference to the first rule, that when 
speaking of tension, that aggregate tension is never meant unless so 
specified. A belt 6" wide, with the same tension, or as taut as a belt 
1" wide, would have 6 times the aggregate tension of the 1" belt. Or 
it would take 6 times the force to slip the 6" belt as it would the 1". 

I prefer to make the readers of this, practical students. I want them 
to learn for themselves. Information obtained in that way is far more 
valuable, and liable to last much longer. In order that the reader 
may more fully understand whether or not a large pulley will hold 
better than a small one ; let him provide a short, stout shaft, say 3 or 
4' long and 2" in diameter. To this shaft firmly fasten a pulley, say 
12" in diameter, or any other size small pulley that may be convenient. 
The shaft must be then raised a few feet from the floor and firmly 
fastened, either in vices or by some other means, so that it will not 
turn. It would be better, of course, to have a smooth-faced iron 
pulley, as such are most generally used. So far as the experiment is 
concerned, it would make no difference what kind of a pulley was 
used, provided all the pulleys experimented with be of the same 
kind, and have the same kind of face finish. When the shaft and 
pulleys are fixed in place, procure a new leather belt and throw it 
over the pulley. To one end of the belt attach a weight, equal say 
to 40 lbs.—or heavier, if desired—for each inch in width of belt 
used ; let the weight rest on the floor. To the other end of the belt 
attach another weight, and keep adding to it until the belt slips and 
raises the first weight from the floor. After the experimenter is 
satisfied with playing with the 12" pulley, he can take it off the shaft 
and put on a 24", a 36", or any other size he may wish, or, what is 



68 


engineers’ manuad. 


ill 5 til ft (0. 

Worcester, Mass. 

MANUFACTURERS OF PURE OAK TANNED 

LEATHER 

BELTING 

Round Belts—Lace Leather—Strappings. 


REPAIRING AND MAKING ENDLESS A SPE¬ 
CIALTY. 

Special Belts Made to Order. 


Dealers in 

RUBBER BELTING, 

EMERY WHEELS, 

MOGUL BEARING METAL, 

HOSE AND PACKING. 

MILL SUPPLIES. 


SEATTLE BRANCH : 119 JACKSON ST. 

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engineers’ manual. 


69 

better, he can have all on the same shaft at the same time. The belt 
can then be thrown over the large pulley, and the experiment 
repeated. It will then be found, if pulley faces are alike, that the 
weight which slipped the belt on the small pulley will also slip it on 
the large one. The method shows the adhesion of a belt with 180° 
contact, but as the contact varies greatly in practice, it is well enough 
to understand what may be accomplished with other arcs of contact. 
But, after all, many are probably at a loss how to account for some 
observations previously made. They have noticed that when a belt 
at actual work slipped, an increase in the size (diameter) ofthe pulleys 
always remedied the difficulty and prevented the slipping. A belt 
has been known to refuse to do the work allotted to it, and continue 
to slip over pulleys 2' in diameter, but from the moment the pulleys 
were changed to 3' in diameter there was no further trouble. These 
observed facts seem to be at variance with and to contradict the 
results of the experiments that have been made. All, however, may 
rest assured that it is only apparent, not real. The resistance to slip¬ 
page is simply a unit of useful effect, or that which can be converted 
into useful effect. The magnitude of the unit is in proportion to the 
tension of the belt. The sum total of useful effect depends upon the 
number of times the unit is multiplied. A belt 6" wide and having a 
tension equal to 40 lbs. per inch in width, and travelling at the rate of 
T per minute, will raise a weight of 240 lbs. l'high per minute. If the 
speed of the belt be increased to 136.5' per minute, it will raise a 
weight of 33,000 lbs. per minute, or be transmitting i-horse power. 
The unit of power transmitted by a belt is rather more than its ten¬ 
sion, but to take it at its measured tension is at all times safe, and 40 
to 45 lbs. of a continuous working strain is as much, perhaps, as a 
single belt should be subjected to. A little reflection will now con¬ 
vince the reader that a belt transmits power in proportion to its lineal 
speed, without reference to the diameter of the pulleys. Having 
arrived at that conclusion, it is then easy to understand why it is 
that a belt working over a 36" pulley will do its work easy, when it 
refused to do it and slipped on 24" pulleys. If the belt travelled 800' 
per minute on the 24" pulleys, on the 36" pulleys it would travel 1,200', 
thus givingit one-half more transmitting power ; if in the first instance 
it was able to transmit but 8 horse power, in the second instance 
it will transmit 12-horse power. All of which is due to the increase 
in the speed of the belt and not to the increase in the size of 
the pulleys ; because, as has been shown, the co-efficient of friction 
or resistance to slippage, is the same on all pulleys with the same arc 
of belt contact. There is no occasion for elaborate and perplexing 
formulas and intricate rules. They serve no useful purpose, but tend 
only to mystify and puzzle the brain of all who are not familiar with 
the higher branches of mathematics ; and it is the fewest number of 
our every-day practical mechanics who are so familiar. In all, or 
nearly all treatises on belting, the writer will tell you that at 600, 800 
or 1000' per minute, as the case may be, a belt 1" wide will transmit 
i-horse power; and yet when we come to apply their rules to prac¬ 
tice, no such results can be obtained one time in ten. The rules are 
just as liable to make the belt travel 400, 1000 or 1600 per minute per 
horse power, as the number of feet they may give as indicating a 
horse power. I have adopted, and all my calculations are based upon 


engineers’ manual. 


70 

the assumption that a belt travelling 800' per minute, and running 
over pulleys both of which are the same diameters, will easily 
transmit 1-horse power for each inch in width of belt. A belt 
under such circumstances would have 180° of contact on both pulleys 
without the interposition of idlers or tighteners. The last pro¬ 
position being accepted as true, and the basis correct, the whole 
matter resolves itself into a very simple problem, so far as a 
belt with 180 0 contact is concerned. It is simply this: If a belt 
travelling 800' per minute transmit one-horse power at 1,600', it will 
transmit two-horse power, or, if 2,400', three-horse power, and so on. 
It is no trouble for anyone to understand that, if he understands 
simple multiplication or division. It is not, however, always the case 
that both pulleys are the same size ; and as soon as the relative sizes 
of the pulleys change, the transmitting power of the belt changes, 
and that is the reason why no general rule has ever or ever will be 
made for ascertaining the transmitting capacity of belts under all 
circumstances. When the pulleys differ in size, the larger of the two 
is lost sight of—it no longer figures in the calculations—the small 
pulley only must be considered. To get at it, the number of degrees 
of belt contact on the small pulley must be ascertained as nearly as 
possible, and used for a guide for getting at the transmitting power, 
the next established basis below. Of course the experimenter can 
make a rule for every degree of variation ; but it would require a 
great many, and is not necessary. I use five divisions, as follows l 
For 180 0 useful effect. 1.00 


“ I57F “ . • -9 2 

“ 13s 0 “ .. 8 4 

“ 112J 0 “ .. . .76 

“ 90° “ ..64 

The experimenters may find that my figures are under-obtained 
results, which is exactly what they are intended to be, more 
especially at the 90° basis. I wish to make ample allowance. To 
ascertain the power a belt will transmit under the first-named con¬ 
ditions, divide the speed of the belt in feet per minute by 800, mul¬ 
tiply by its width in inches and by 100. For the second, divide by 
800, multiply by width in inches and by . 92. Third place, divide by 800, 
multiply by width in inches and by .84. Fourth place, divide by 800, 
multiply by width in inches and by .76. Fifth place, divide by 800, 
multiply by width in inches and by .64. As an example : What 
would be the transmitting power of a 16" belt, travelling 2,500' per 
minute by each of the above rules? 

1st. 25004-800 = 3.125 x 16 and 1.00 = 50 h. p. 

2nd. 25004-800 = 3.125 x 16 and .92=46 h. p. 

3rd. 25004-800 = 3.125 x 16 and .84 = 42 h. p. 

4th. 25004-800 = 3.125 x 16 and .76 = 38 h. p. 

5th. 25004-800=3.125 x 16 and .64 = 32 h. p. 

As I have said, if the degrees of contact come between the divisions 
named above, in order to be on the safe side calculate from the first 
rule below it, or make an approximate, as they like. If the above 
lesson is studied well and strictly used, there can be no excuse for 
any mechanic putting in a belt too small for t{ie work it has to do, 
provided he knows how much there is to do, which he ought, 
somewhere near at least. 








engineers’ manual. 

STEAM. 


71 


Sensible and Latent Heat. Heat given to a substance, and warm¬ 
ing it, is said to be sensible in the substance. Heat given to a sub¬ 
stance and not warming it is said to become latent. —Sir Wm. 
Thomson. 

Latent Heat is the quantity of heat which must be communicated 
to unit mass of a body in a given state, in order to convert it into 
another state without changing its temperature.— Maxwell. 

If 1 lb. of ice be placed where it will receive heat uniformly at the 
rate of 18 units per minute it will melt gradually so that more and 
more of it becomes water until at the end of 8 minutes it is all melted; 
during this time the temperature of the ice and water will have re¬ 
mained at 32 0 . This shows that 144 units of heat have been spent 
without increasing the sensible heat ot the substance ; these 144 heat 
units have become latent in the water and this is the amount that 
must be taken from 1 lb. of water at 32 0 to change it to ice at 32 0 . 

If the heat is still kept on the 1 lb. of water at 32 0 , it will in 10 
minutes receive 10 x 18=180 units, which will raise its temperature to 
212 0 when it will boil. It will continue to boil and become gradually 
converted into steam until in 53.7 minutes the whole is so changed. 
During this change its temperature will remain steadily at 212 0 and 
therefore 53.7 x 18=966.6 units will have becomeiatent in steam. 

Latent Heat of Fusion. When a body passes from the solid to 
the liquid state, its temperature remains nearly stationary, at a cer¬ 
tain melting point during the operation of melting; and in order to 
make that operation go on, a quantity of heat must be transferred to 
the substance melted. This quantity is called the latent heat of fusion. 
In ice this is 144 units. 

Latent Heat of Evaporation. When a body passes from the solid 
or liquid to the gaseous state, its temperature during the operation 
remains stationary at a certain boiling point, depending on the pres¬ 
sure of the vapor produced, and in order to make the evaporation go 
on, a quantity of heat must be transferred to the substance evaporated, 
This heat does not raise the temperature, but disappears in causing 
it to assume the gaseous state, and is called the latent heat of Evap¬ 
oration. 

Total Heat of Evaporation is the sum of the sensible and latent 
heats of evaporation. To raise 1 lb. of water from freezing point 
(32°F.) to the temperature of evaporation (32°F.) takes 180° sensible 
heat units and the additional heat required to evaporate it is called 
the latent heat; to evaporate 1 lb. water at 212 0 into steam at the 
same temperature takes 966.6 heat units. The total heat of evapor¬ 
ation for water is therefore = 180 +- 966 6= 1146.6. 

If steam is generated at a higher temperature than 2i2°F., the 
sensible heat increases, and the latent heat decreases. 

To find the latent heat of steam for any temperature, the follow¬ 
ing formula will be found very nearly correct : 

Latent heat = 966.6 - . y(t - 212 0 ) 
where t ■= the temperature of evaporation, 


72 


engineers’ manual. 


\ 

From this we see that since the temperature of the steam is 
raised, the latent heat diminishes only . 7 of the increase in the sensible 
heat it is therefore obvious that the total heat increases. For all tem¬ 
peratures aoove 212 0 the latent heat is less than 966.6, and for all 
temperatures below 212 0 th* latent heat is greater than 966.6. 

Example -What is the latent heat of steam when the thermo¬ 
meter registers 332°F. ? Ans.—882°. 

When the latent heat is found, at any temperature, the total heat 
of evaporation is very easilv determined. 

Total heat of steam = Sensible + Latent heat 

= (/-32°) +966.6- ‘7(t — 212°) 

= 1083+ .3t 

Example—Find the total heat of steam at 2i2°F. Ans.—1146.6. 


Quantity of Water Required for Condensation. 

Let H — total heat calculated from 32°F. 
t x "= temperature of steam 
t^— temperature of water 
t s — resulting temperature 
X = lbs. of water at t 2 

The loss of heat from the Steam = the gain of heat by the water. 
The heat given up by 1 lb. of steam = H- (t a - 32) 

The heat gained by X lbs. water=AT(^ 3 - 1 2 ) 


A H-(t s - 3 2) = X(t s -t,) 

But H — 1083+ .3t t 
.-. 1083+ .3*1- (t a -32) = X{t a -t i ) 

Y — Ill 5 + - 3*1 -*> 

Example I. — If 1 lb. of steam at 2i2°F. be mixed with X lbs. of 
water at 6o°F. What is the value of X when the resulting tempera- 
ture is ioo°F. ? Ans. — 26.96 lbs. 

Example II. — Steam enters the condenser at a temperature of 
!42°F. to be condensed into water at i20°F ; the circulating water 
enters at 6o°F. and is discharged at ioo°F., find how many lbs. of 
circulating water will be required per lb, of steam. Ans. — 25.9 lbs. 




engineers’ manual. 


73 


TABLE I. 

Properties of Saturated Steam from 32 0 to 212 0 . 


V 


Temper¬ 

ature. 

PRESSURE. 

Temper¬ 

ature. 

PRESSURE. 

Inches of 
Mercury. 

Lbs. per Sq. 
Inch 

Absolute. 

Inches of 
Mercury. — 

Lbs. per Sq. 
Inch 

Absolute. 

32 

.181 

.089 

125 

3-933 

1.932 

35 

.204 

. 100 

130 

4-509 

2.215 

40 

.248 

. 122 

J 35 

5.174 

2.542 

45 

.299 

.147 

140 

5.860 

2.879 

50 

.362 

.178 

M 5 

6.662 

3273 

55 

.426 

.214 

150 

7.548 

3.708 

60 

•517 

•254 

*55 

8-535 

4 193 

65 

.619 

304 

160 

9.630 

* 4-731 

70 

•733 

.360 

165 

10.843 

5 327 

75 

.869 

.427 

170 

12.183 

5 985 

80 

1.024 

•503 

175 

13654 

6.708 

85 

1.205 

•592 

180 

15.291 

7511 

90 

1.410 

•693 

185 

I 7-°44 

8-375 

95 

1 647 

.809 

190 

19.001 

9-335 

100 

1 • 9 I 7 

.942 

195 

21.139 

10.385 

105 

2.229 

1.095 

200 

23.461 

11.526 

no 

2-579 

1.267 

205 

25-994 

12.770 

1 15 

2.976 

1.462 

210 

28.753 

14.126 

120 

3-430 

1.685 

212 

29.922 

14.700 


A strip of looking- glass held behind a glass water gauge makes 
it easier to see the water-line. 

In cutting rubber for gaskets wet the knife often with a strong 
solution of potash. This makes the cutting easier. 

In making rubber joints, chalk the rubber well before screwing 
up the flanges. When this is done the joint will always come apart 
easily. 

To ascertain whether a plate is burned or crystallized, take a 
thin, sharp chisel and cut a thin chip for an inch or two ; if the plate 
is good the chip will curl up. 

The calorific power of wood is about .4 that of the same weight 
of good coal. The fuel value of different woods is praotically the 
same, provided they are equally dry. 

A good quick setting rust joint is formed of sal-ammoniac 
powdered, 1 lb.; flour of sulphur, 2 lbs.; iron borings, 80 lbs.; mix to 
a paste with water. A slow setting rust joint is made up of sal- 
ammoniac, 2 lbs.; sulphur, 1 lb.; iron borings, 200 lbs. This is best, 
if the joint is not needed for use at once. 




























74 


engineers’ manual. 




TABLE II. 

Properties of Saturated Steam. 


(From Peabody’s Tables). 


Press, in lbs. 

per sq. in. 
ab’ve vacu’m 

Temperature 
in degrees 
Fah. 

Total heat 
units from 
water at 32 0 . 

Heat of va¬ 
porization. 
Latent heat 
units. 

Weight of 
cubic foot 
in pounds. 

Volumes 
of one 
pound in 
cub. feet. 

i 

101.99 

1113.1 

1043.0 

.00299 

334-5 

2 

126.27 

1120.5 

1026.1 

.00576 

173.6 

3 

141.62 

1125.1 

IOI 5-3 

.00844 

118.5 

4 

I 53°9 

1128.6 

1007.2 

.01107 

90.33 

5 

162.34 

II 3 I -5 

1000.8 

.01366 

73.21 

6 

170.14 

1133-8 

995 • 2 

.01622 

61.65 

7 

176 90 

1135-9 

990.5 

.01874 

53-39 

8 

182.92 

1137-7 

986.2 

.02125 

47.06 

9 

188.33 

1 139-4 

982.5 

.02374 

42.12 

IO 

193-25 

1140.9 

979.0 

.02621 

38.15 

i 5 

213.03 

1146.9 

965.1 

.03826 

26.14 

20 

227.95 

ii 5 i -5 

954-6 

•05023 

19.91 

25 

240.04 

H 55 -1 

946.0 

.06199 

16.13 

30 

250.27 

1158-3 

938.9 

.07360 

13-59 

35 

259 -19 

1161.0 

932.6 

.08508 

n -75 

40 

267.13 

1163.4 

9 2 7 • 0 

.09644 

10-37 

45 

274.29 

1165.6 

922.0 

•1077 

9.285 

50 

280.85 

1167.6 

9 i 7-4 

.1188 

8.418 

55 

286.89 

1169.4 

9131 

.1299 

7.698 

60 

292.51 

I I 7 I -2 

9 ° 9-3 

.1409 

7.097 

65 

297.77 

I 172.7 

9°5 • 5 

• L 5 I 9 

6-583 

70 

302.7! 

I 174-3 

902.1 

. 1628 

6.143 

75 

307•38 

U 75-7 

898.8 

• 1736 

5.760 

80 

311.80 

H77.O 

895.6 

.1843 

5.426 

85 

316.02 

II78.3 

892.5 

•1951 

5.126 

90 

320.04 

II79.6 

889.6 

.2058 

4.850 

95 

323-89 

1180.7 

886.7 

•2165 

4.619 

100 

327-58 

1181.9 

884.0 

. 2271 

4-403 

!05 

33 i-i 3 

H82.9 

881.3 

.2378 

4.205 

110 

334 • 5 6 

I184.O 

878.8 

. 2484 

4.026 

115 

337-86 

H85.O 

876.3 

.2589 

3.862 

120 

34 i-o 5 

ii86.0 

874.0 

.2695 

3 - 7 II 

125 

344 -13 

1186.9 

871.7 

. 2800 

3 - 57 i 

130 

347.12 

1187.8 

869.4 

.2904 

3-444 

I40 

352-85 

1189-5 

865.1 

•3113 

3.212 

150 

358.26 

1191.2 

861.2 

•3321 

3.011 

l6o 

363-40 

1192.8 

857 • 4 

•3530 

2-833 

170 

368.29 

1i 94-3 

853-8 

■3737 

2.676 

l8o 

372-97 

H 95-7 

850-3 

• 3945 

2-535 

190 

377-44 

11971 

847.0 

•4153 

2.408 

200 

38 i -73 

1198.4 

843.8 

•4359 

2.294 































engineers’ manual. 


75 


Expansion of Steam. 

When saturated steam expands in a non-conducting cylinder, 
and during its expansion performs mechanical work, its pressure 
falls on account of increase of volume and because of liquefaction. 
Rankine’s approximate rule for the relation between pressure and 
volume, expanding under the above conditions is “ The pressure varies 
nearly as the reciprocal of the tenth power of the ninth root of the space 
occupied that is, 

^>=pressure and z;=volume ; 
then, px— V or px v i_ 0 

or, pv V°~constant. 

This curve is very nearly an adiabatic curve, and falls considerably 
below the hyperbolic or isothermal curve. 

Although the above is useful in certain theoretical investigations, 
it is of little practical use, because non-conducting and non-radiating 
cylinders do not exist. 

In steam engines fitted with good steam jackets, in which steam 
enters in a moist condition, a .considerable quantity of the heat passes 
from the jacket to the steam in the cylinder. When this quantity of 
heat is sufficient, not only to do the work performed by the steam, 
but also to convert a portion of the wet steam into dry saturated 
steam during the expansion, the relation between pressure and 
volume is approximately expressed by Boyle’s law, viz.: 

pressure X volume=constant, 

and the curve is an hyperbola. The hyperbolic curve is usually 
adopted for rough calculations in expansion. Boyle’s law may be 
briefly stated as follows : The pressure of a portion of gas at a con¬ 
stant temperature varies inversely as the space it occupies. 

The following examples will show clearly the application of 
Boyle’s Law : 

I. —Back pressure 4 lbs., steam pressure 30 lbs., clearance How 

far must piston be from end of its stroke at compression to 
compress the enclosed vapor in cylinder, so that the pressure 
shall rise to that of the steam in the steam chest. 

Steam in the clearance space has a volume of at 30 lbs. 
pressure. The back pressure has to increase from 4 lbs. to 
30 lbs.; therefore, the volume has to decrease from 30 to 4, or 
T 2 -. At compression the steam occupies \° of |"=3f". From 
3I" deduct the amount of clearance, and we get 3^' as the 
distance the piston is from the end of stroke where 
compression began. 

II. —Steam pressure 45 lb. gauge, cut off at £ lb. stroke. Find the 

mean pressure. 

Method— Draw a horizontal line A B to represent the length of 
the stroke and also the line of volumes, divide this line into 6 equal 



76 


engineers’ manual. 


PATENTS, 

DESIGNS, 

TRADE-MARKS, 

COPYRIGHTS, 

CAVEATS. 


SUPERIOR 
SERVICE 
IN ALL 
BRANCHES 
OF THE 
WORK 


PATENTS 


Mechanical Drawing 


PIERRE BARNES & CO. 

REGISTERED PATENT ATTORNEYS 


WRITE FOR OUR BOOK ON PATENTS. 


74 and 75 STARR-BOYD BLOCK, 
SEATTLE, WASH. 






engineers’ manual. 


77 

parts. From A erect a perpendicular A C to scale representing line 



of pressures. In all steam calculations the pressure must be taken 
from zero or absolute, therefore 45 lbs. gauge pressure is equal 60 lbs. 
absolute. The line A C will represent 60 lbs., and as cut off does not 
take place until the piston has travelled % of the stroke we will have 
the same pressure all the way along from C to D. At D the valve is 
closed and the rest of the work done in the cylinder is by expansion. 
At the point F which is § or £ of the stroke, the volume of the steam 
has increased to twice its original amount, therefore its pressure will 
be £ or 30 lbs. At G the volume is increased to 3 times the original 
volume, therefore its pressure is only ^ or 20 lbs. At H volume is 4 
times pressure = 15 lbs., and at J the pressure is 12 lbs., correspond¬ 
ing to 5 times the original volume ; and at the end of the stroke the 
piston has moved over 6 times the distance A E, therefore pressure is 
only ^ of the original = ^ of 60 lbs. = 10 lbs. 

The mean pressure is obtained by taking the sum of the average 
pressures between the points of division on the diagram, and the 
dividers by the number of divisions, as follows : the average pressure 

60 + 30 

between A and E is 60 lbs., between E and F = —-— == 45, and so 

on, and taking the sum of these and dividing by 6 we get 28.66 lbs. 
as the mean pressure. 

Suppose this engine had £ of the cylinder volume of clearance. 
(This is an excessive amount, but it is to show the effects of clearance 
to a somewhat enlarged extent on the diagram). 

Proceed as before, and draw A B to represent the stroke of the 
engine + i of the stroke, or A B = | of the stroke of engine. Divide 
A B into 7 parts. The first of these will represeut the amount of 
clearance to the same scale as the stroke, and the distance A E will 



represent the amount of steam there is when piston has travelled £ of 
its stroke or to the point when valve has just closed. When piston has 


















78 


engineers’ manual. 


V 


travelled to i^the volume has increased from 2 to 3, therefore pres¬ 
sure has decreased from 60 to 40. The pressure at G is 30 lbs., be¬ 
cause volume has doubled, and so on, when we arrive at the end of 
the stroke with a pressure of 17.1 lbs. By taking the means of these 
and adding them, and then dividing by 7 we get an average pressure 
of nearly 39 lbs. 

Ratio of Expansion. 

The ratio of expansion as usually understood is the ratio of the 
cylinder volume to that of the volume of the cylinder at point of cut¬ 
off, or the ratio of the length of the stroke to that part of the stroke 
travelled by the piston up to the point of cut-off, or 

. cylinder volume 

Ratio of Expansion= — ----— 

volume to point of cut-off 

_ length of stroke 

length of stroke to point of cut-off 

If clearance is taken into account the true or actual ratio of ex¬ 
pansion is much less than the ratio given above. 

, , ,. c . cylinder volume + clearance 

the actual ratio of expansion =— - -- --= 

volume to cut-off + clearance 


No. of volumes to which the initial volume is expanded. 

Example—Stroke 4 feet; clearance ; cut off at £ stroke. Find 
ratio of expansion (1) without clearance, (2) with clearauce. Ans.— 
4 5 3 b 

Expansion of Steam. 


Let L — 

4 - _ 


C = 
P = 

P — 
R = 
H = 
K — 


length of stroke in inches. 

distance travelled by the piston before steam is 
in inches. 

clearance in inches. 

initial absolute pressure. 

mean pressure during stroke, in lbs. 

, , . L +c 

actual rates of expansion = 
hyp. log. of R. 
i+H 
R 


t+C 


cut off, 


To Find the Mean Pressure— 


p f t+hyp - 
L > R J 

= =P .K. 

To Find the Initial Pressure— 


p=jL= p 

To Find the hyp. log. of R — 

h = 

P 


(1) 


( 2 ) 


( 3 ) 











engineers' manual. 


79 


To Find the Ratio of Expansion— 

R = p[ l+I D. (4) 

P 

The values of R y H and K can be readily found in lie following- 
table when the point of cut-off is known : 


Cut-off. 

Ratio of 
Expansion. 

Hyper, 
log. R. 

K or 

1 + hyp. log. 
R 

3V 

TO 

2.302 

•3302 

TV 

9-5 

2.251 

.3422 

b 

9 

2 < I 97 

•3552 

& 

8-5 . 

2.140 

•3694 

1 

8 

2.079 

•3849 

T5 

7-5 

2 . 01 5 

.4020 

\ 

7 

I.946 

.4208 

2 

VS" 

6-5 

I.872 

.4418 

} 

6 

I. 791 

•4653 

If 

5-5 

I -705 

.4917 

i 

5 

I .609 

•5219 

1 

4-5 

i • 5°4 

•5564 

1 

T 

4 

1.386 

■5965 

f 

3-5 

1.252 

.6438 

| 

3 

1.098 

.6962 


2-5 

.916 

.7666 

\ 

2 

• 6 93 

.8465 

2 

T 

i -5 

•405 

• 937 ° 

I 

1 

.000 

i.0000 


Note—From the results obtained by the above rules, the back 
pressure has to be deducted. 

The following examples will show the method of working the 
various formulae : 


Example*!.—Initial pressure 120 lbs. absolute. Cut-off \ stroke. 
Back pressure 21.58 lbs. Find mean effective pressure. M. E. P.= 
mean forward pressure,—mean backward pressure. By Formula 
(1) we get 

M. E. P. = 120 f 1 + h yP-l°g- -[ _ 2I . 5 S 
^ 4 J 

1 + 1. 

4 

Example II.—The M. E. P., as measured on a diagram, is 56 lbs. 
The scale of the diagram is and the back pressure line is ^ of an 
inch above the atmospheric line. If cut-off takes place at ^th stroke, 
find the initial gauge pressure. By Formula (2), 

p—P_ = S^ + b of 4 °+ * 5 —ib St absolute 
k .5219 

or 131 lbs. guage. 


= 120 


—21.58=50 lbs. 
















8o 


engineers’ manual. 


Example III.—Find the hyp. log - , when initial pressure is 120 lbs. 
absolute ; cut-off at \ stroke ; mean forward pressure 71.58 lbs. By 
Formula (3), 

B=t* - 1 = 7 LS 8 x 4 _ i= ,. 3 86 
P 120 

To Find the Mean Pressure by the above table— 

Rule : Find the value of K corresponding to the cut-off or ratio 
of expansion. Multiply this by the initial absolute steam pressure, 
and from the product subtract the back pressure. 

Example—The pressure as indicated by gauge is 100 lbs. Cut¬ 
off takes place at f stroke. The engine is exhausting at 5 lbs. above 
the atmosphere. Find the M. E. P. 

100 lbs. gauge=ii5 lbs. absolute: 5 lbs. gauge=20 lbs. absolute. 
In table, the value of K agreeing to cut-off at f stroke is .6962 .'. 
.6962 x 115—20=60 lbs. 

Graphic Method of Finding the Mean Pressure. 

From the centre A at the distance A C describe the arc C E D. 
From A measure off A B as distance equal to i of A C. Join B D. 
This line B D represents the stroke of the engine and D represents 
the beginning of the stroke. From D measure off the distances as 
indicated by the ordinates f, f, etc., corresponding to f, A, etc., 
of the stroke. For instance : DF is equal to i of the stroke, or 
D F 

Make B C by construction = 2". Then the ratio of the 

mean pressure to that of the initial absolute pressure is equal to the 
length of the ordinate corresponding to the point of out-off divided bv 
the length BC. y 

Example—The ordinate EF we find is if, then if-f 2= the ratio 
of the mean to the initial abs. pressure. If initial pressure was 100 
lbs. then i|-t-2 x 100=69 lbs. mean pressure. 



The following table gives the approximate length of the ordinates 






























engineers’ manual. 


81 


in 32nds of an inch, together with the exact length in decimals : 


Cut- 

Ratio of 

Approx, length 

Exact length 

Ratio of ordin¬ 

off. 

Expansion. 

of ordinate. 

of ordinate 

ate to B C. 

i 

8 

ft 

•77 

.385 

i 

6 

tt 

.9306 

•4653 

1 

T 

5 

f! 

1.044 

.522 

1 

T 

4 

3u 

1.1930 

•5965 

1 

IF 

3 

44 

TTS 

1.3924 

.6962 

6* 

2-5 


1 -533 2 

.7666 

i 

2 


1.6930 

.8465 

3 

i-7 

if4 

1.7820 

.891 

t 

1 -5 

jff 

1.8740 

•937 

¥ 

i-33 

If® 

1.9200 

.960 


The Rule for finding the mean pressure by the above method is : 

Divide the ordinate corresponding to the ratio of expansion by 
the length B C and multiply by the initial pressure. 

Example The stroke of an engine is 24" and steam is cut off 
when piston has travelled 6". Find the mean pressure if the initial 
pressure is 115 lbs. absolute. 

24 

Ratio of expansion = -^-=4. By table we get the length of 

the ordinate to be 1.1930, and this divided by 2 gives us .5965. This 
multiplied by 115 gives us the mean pressure = 68.6 lbs. 

CARE OF STEAM BOILERS. 

The management of Steam Boilers in all establishments is a 
subject of great importance, and one which does not, in many cases, 
receive the care and attention necessary in order to obtain the most 
economical results. When a steam user has decided to purchase a 
boiler, he should take steps to determine exactly what size and style 
will best suit his requirements. 

If his engineer has the necessary ability, he should be requested 
to make an evaporative test of the plant, and also to indicate 
the engine in order to determine the exact amount of water required 
to be evaporated, and the number of horse power exerted by the 
engine. 

Having this information, it is an easy matter to arrive at the 
dimensions of the boiler required ; but it must be remembered that 
it is always in the interest of economy to have a boiler larger than is 
necessary for the actual requirements, as the grate surface can be 
proportioned to suit the case, fires can be run without any forcing, 
and good combustion and consequent economy of fuel secured, to say 
nothing of the increased length of the life of boilers used under 
these conditions. If the engineer has not the ability necessary to 
determine these points, then some engineer of ability and good 
















82 


engineers’ manual. 


I 

y 

standing- snould be entrusted with the getting up of specifications'^ 
both for the building of boiler and brickwork The boiler should be 
inspected frequently during construction, and, when completed, it 
should be thoroughly inspected and tested to one and one-half times 
the pressure it is desired to work, by hydrostatic pressure. 

After the boiler has been set in position and the brickwork com¬ 
pleted, it should be allowed to stand, if possible, for a week in order 
to give the brickwork a chance to dry and set. After this, the boiler 
may be filled to the proper level and a small fire kept burning under 
it for a few days before being put to work, great care being used so 
as not to heat up the boiler and brickwork too quickly. 

In starting up a new boiler, it is a good plan to put in a few lbs 
of sal. soda with the water, and then, after brickwork is well dried 
and set, to let down fire and steam, run off the water and give the 
boiler a good washing out. This treatment will be found to prevent 
the foaming which so often happens when starting up a new boiler, 
and is caused by the grease left in it by boilermakers. 

From the time a boiler is started to work certain influences are 
at work, which, if left to themselves, will materially shorten its term 
of usefulness and safety; and it is the duty of the engineer to use 
every effort to check and counteract them. 

The importance of the duties of the engineer and fireman are 
not as fully understood by many of our steam users as they should 
be, and too many owners are inclined to think that everything is all 
right as long as the machinery keeps on the move. A good, intel¬ 
ligent, painstaking and thinking engineer or fireman, compared 
with the careless and indifferent man, will save his wages several 
times over. It is a well-known fact to many firms who have given 
the matter attention that a good fireman is almost invaluable, and 
that the difference in the fuel bill between a really good fireman and 
an indifferent man is astonishing at the end of a year. 

The fireman should at all times, before starting his fire, see that 
the water in boiler is at proper level. He should not be satisfied by 
merely looking at the water-glass, but should open the cock at bottom 
of glass, and also try the gauge cocks. Many accidents have 
occurred by neglecting this duty. 

When sure that the water is all right, he should see that blow-off 
cock is in order and closed, that the ashpit is clear of ashes, that the 
tubes are clean, and that the safety valve is raised off its seat, or 
that some valve or cock is open to the atmosphere until steam issues 
from it. The grate bars should now be covered (with coal) from the 
bridge wall toward the furnace door for about 3 feet, and should 
then put in some light wood on the grate in front of the coal, and 
with a little oily waste set fire to it. 

When the fire has taken well hold of the wood a little coal may 
be put on it. During this time the ashpit should be closed and the 
furnace door left open a little in order that the flames may be com¬ 
municated to the coal at the back of furnace. 

As soon as a good fire is burning in the front of furnace, it may . 
be pushed back a little and the ashpit damper opened. The fire 
should not be forced, but should be allowed to work up gradually, 




engineers’ manual. 


83 


as the unequal strains some boilers are subjected to through forcing 
the fire when boiler is cold have caused leakage, and made expen¬ 
sive repairs necessary. In boilers of the Galloway, Lancashire and 
Cornish type, it is necessary to use great care in firing up from cold 
water, owing to the temperature of the water in the lower part of 
shell remaining low for a considerable length of time. The fires 
should be maintained level and of a uniform thickness, but the 
thickness must be determined by the demand for steam, condition 
of the chimney draft, and quality and nature of the fuel. 

The firing is best done when the combustion in furnace is good, 
and consequently but little dense smoke is given off. Dark spots in 
the fire, abundance of smoke, unsteady steam pressure, unsteady 
water line, dirty tubes, coal in ash heap, are all evidences of careless 
firing, and should not be tolerated. Experience is the only thing that 
will prove the best methods of handling the different kinds of fuel 
under the different conditions to be met with in practice. 

In the boiler room there should be a place for everything, and 
everything should be kept in its place. 

All the fittings, mountings, boiler front, etc., should be kept clean 
and free from leaks. 

The coal should be put on fire at regular intervals and lightly. 
If the furnace is large, it may be advisable to coke the fire, i.e., to 
fire the green coal in front of furnace and allow the smoke to pass 
over a bed of incandescent, full at the back, and be consumed ; then 
push it back and add more coal in front. 

Sometimes side firing works very well; i.e., to always have one 
side of the fire incandescent when firing green coal on the opposite 
side. But no hard-and-fast rule can be set for every condition, and 
much must be left to the judgment of the fireman in each individual 
case. When firing or cleaning fires, where the chimney draft is very 
strong, it is advisable to check the stack damper to prevent too great 
a quantity of cold air entering the furnace and causing undue con¬ 
traction of the plates. In boilers having large furnace, it is well 
when cleaning fires to clean one side at a time. 

The fires should be banked at night, as it is more economical 
than to allow fires to burn out and re-light them in the morning, and 
it also saves the life of boilers to a certain extent, as, when fires are 
banked, the boiler is not subjected to so many strains by expansion 
and contraction. 

The feed water should be kept constantly on, and the water-line 
maintained at the proper level all the time. Every day the steam 
pressure should be raised to the blowing-off point, so that the fire¬ 
man may know that the safety valve is in working order. If at any 
time, from any cause, the gauge should show the pressure increasing 
rapidly up to or past the limit, the feed should at once be put on, 
draft checked, and in some cases it may be necessary to open the 
furnace doors. Should the water in boiler at any time get danger¬ 
ously low, then close dampers and open smoke-box doors immediately, 
and cover fires with damp ashes, or, if there are none at hand, small 
green coal may be used. Do not put on the feed, but allow the 
boiler to cool down some. After this the feed may be put on, and 


engineers’ manual. 


84 

the tubes at back end examined for fear they may have been caused 
to leak from overheating. 

If the water-gauge glass and try cocks are attached to a 
column, there should be a blow-off pipe from bottom of column of at 
least diameter, and this pipe should be carried to main blow-off 
pipe or sewer, and should be blown off at least once every two hours. 

In cases of foaming or priming, if not caused by faulty construc¬ 
tion of boiler, it can usually be prevented by putting on more feed 
and opening blow-off, thus changing the water in the boiler. But if 
the foaming is very violent, it may be necessary (in order to deter¬ 
mine the water-level in boiler) to close, or partially close, the engine 
throttle, open the furnace-door and increase the feed, and blow off 
the boiler a little at intervals. A surface blow-off cock is a good 
thing when a boiler foams, as by its use the scum and dirt can be 
cleared off surface of water. 

A boiler should be cleaned out at regular intervals, but the 
length of time between such cleanings must be determined according 
to the nature of the feed water. A boiler using feed water from the 
Lake may be run for from six to eight weeks before cleaning, while 
on the other hand using feed water from a small stream it may be 
necessary in the spring of the year (when the water is very dirty) to 
clean the boiler every week. 

When about to clean and wash out the boiler, the brickwork 
should be allowed to cool down as much as possible before the water 
is run off; then the hand hole covers should be taken out, and all 
mud and deposit removed by scraping out, then the hose should be 
used with a good water pressure and boiler washed out thoroughly. 
After this has been done the water should be all drained out of 
bottom of boiler, and a light put into it through hand hole to make 
sure that no scale or mud remains on the bottom. 

The manhole should be taken out once every three months, when 
the fireman should go inside, and, with proper cleaning tools, scrape 
off all deposit and dislodge all accumulations of scale, which will 
fall to the bottom of shell and can be removed through the hand 
holes. After this has been done thoroughly, the boiler may be washed 
out well through the manhole. All joints should then be made, 
care being taken to make them perfectly tight, as, if allowed to leak 
and run down the boiler, it will cause corrosion of plates, and in time 
necessitate repairs. All soot and ashes should be removed from 
under boiler previous to commencing to wash out, and tube ends and 
bottom of boiler, seams, etc., should be carefully examined for leaks, 
and if any are found they should be caulked and made tight without 
delay, as, if left for any length of time, they will cause expensive 
repairs and delays. If the boiler is subject to inspection, the bottom 
of shell should be swept off, all dust and ashes removed from flues, 
and every facility ^iven the inspector to enable him to do his work 
thoroughly. 

A man in charge of a steam boiler should have a due sense of 
his responsibilities. He should be cool and collected in case of 
emergency, sober and industrious at all times, and should never put 
off till to-morrow the things th^t ought to be done to-day. This 


engineers’ manual. 


85 


article on Care of Steam Boilers is not written for experienced 
engineers, but rather for the young fireman who is seeking informa¬ 
tion, and who has a desire to advance in his chosen calling. If it is 
read by even a few of the latter and proves in any way beneficial to 
them, then one of the objects in publishing this book will have been 
accomplished. 

BOILER SETTINGS. 

The brick work about a boiler should be thick to prevent loss by 
radiation—a 21" wall should be used if possible. All flues and surfaces 
exposed to action of heat should be lined with the best fire brick. 

It is not a good plan to convey gases back over top of boiler, 
unless there is space enough for a man to enter and clear off soot. 

The distance from grate bars to lower portion of boiler shell 
should not be less than 24" ; 26" and 28" are not too great, and in large 
shells 30" can be employed. 

The bridgewall should curve to conform with the boiler shell. Ten 
(10) inches makes a good space between wall and shell. Back of 
bridge wall the surface should be paved with hard brick, the surface 
dipping down to a depth at the rear end of boiler of about 18" to 24", 
according to size of shell. The distance between back tubes, shell 
and back wall should be 18" for a 48" shell, and 24" for a 72" shell. 

Boiler walls will crack, and no form of construction seems to 
entirely prevent this. Walls with air spaces are as liable as those 
without, with the danger of leaking more air when they do crack. 

The best method to hold boiler walls together is with “ brick- 
staves” or “brick-bars.” The best form is railway iron with ends 
mashed down under a hammer to allow for drilling for tie rod. Most 
builders do not supply “ brick-staves” unless specially ordered. 

The cheapest form of fire-front is the so-called “ half-arch,” 
which does not cover any more of the front of the furnance than is 
absolutely decent. On small boilers it is employed as a support. 
For a good job a “full flush front” should be used, with damper 
plate and damper. 

Boilers, now-a-days, are not set in batteries, all to work together 
as a unit. They are, and should be, set so that each boiler is inde¬ 
pendent of the others in the battery. In this way, any one can be 
shut down for cleaning and repairs. This arrangement does away 
with the old-fashioned steam and mud-drums, which connected the 
boilers of the battery together. Do not buy either a mud-drum or 
steam drum—they are a source of trouble, danger and expense. 

EVAPORATIVE TESTS. 

It is important to owners of steam plants that they should some¬ 
times take steps to determine whether the efficiency of the plant is 
up to the standard ; or, in other words, to determine whether or not 
the fuel which is being consumed under the boilers is evaporating as 
much water as is possible. The heat value of coal varies considerably, 
and it is very seldom that this fact is taken notice of by engineers in 
making evaporative tests. There are three different methods of 


86 


ENGINEERS’ manual. 


determining - the caloric value of fuels, viz.: by chemical analysis, by 
use of the calorimeter, and by the actual measurement of water eva¬ 
porated per pound of fuel consumed in the furnace. The first process 
is of course impossible for an engineer to accomplish, and would 
require the services of an analytical chemist, and even then the result 
would be only approximate. The second method is probably more 
satisfactory, and its operation is as follows : A sample of the fuel to 
be tested, mixed with chlorate of potassium, is placed in a copper 
vessel with an open mouth, and this is submerged mouth downwards 
in water of a known quantity. Combustion then takes place and the 
heat value of the coal is determined from the rise in temperature of 
the water. If the second method be used to determine the value of a 
fuel, and in order to secure fairly accurate results, it is necessary to 
test a large number of samples taken from different parts of a pile, so 
as to ensure average results. In most of the evaporative tests made 
no thought is given to the heat value of the fuel, yet the qualities of 
coal vary just as much as other articles of commerce. 

The third and most practical method of determining the value of 
coal is to test it for evaporative duty under a clean, well designed and 
well set boiler, for it matters but little what may be the value of a 
fuel according to the analysis, or the calorimeter test, when we can 
only obtain certain results from it when consumed in the furnace of a 
steam boiler, and it is by this test that we must determine the value 
of our fuel and the efficiency of our steam boilers and engines. 

To rightly determine the heat value of any fuel (for comparative 
purposes and to do justice to the fuel) it is necessary that all condi¬ 
tions as.regards style of boiler, setting, draft, ratio of grate to heat¬ 
ing surface and skill in handling the fuel shall be the same, therefore 
it would be most unjust to condemn a fuel because in a test at Messrs. 
A. & Co. only 6 \ lbs. of water were evaporated per lb. of coal, be¬ 
cause a test might be made at Messrs. B. & Co. using the same fuel 
and result in showing an evaporation of 8^ to 9 lbs. of water per lb. 
of coal. 

It may be misleading to judge of the value of a fuel on the 
strength of a test made at So and So’s establishment, and equally so 
to condemn a boiler on the report of an evaporative test, as it is ab¬ 
solutely necessary to know all the conditions under which the test 
was made before reaching any conclusion as to the value of the fuel 
or the efficiency of the boiler. 

It is interesting to notice the different grades of combustion 
attained in our industrial establishments, as indicated by the output 
of smoke from the chimneys. In some places we see vast volumes 
of black smoke rolling away, and in others just a light smoke is 
noticeable, and it only for a few moments after charging the furnace 
with green coal. In many instances, where a large quantity of 
smoke is sent off from the chimney, it could be very much reduced 
by careful stoking and some knowledge of the laws of combustion. 

Whenever large quantities of smoke issue from the chimney, we 
know there is a waste of fuel through poor combustion ; conse¬ 
quently, the evaporative duty of the coal will be less than it should 
be, according as the combustion is good, bad or indifferent. 


engineers’ manual. 


87 


We know we can never utilize all the heat generated by com¬ 
bustion by transferring it to the water in the boiler, for the reason 
that the gases from furnace cannot be reduced lower than the 
temperature of steam and water within the boiler, and, in addition 
to this, a certain amount of heat is necessary over and above the 
temperature of the atmosphere to induce draft in the chimney so as 
to induce the necessary amount of air for combustion to enter the 
furnace through the burning fuel. 

It is possible, by proper proportioning of grate surface to heating 
surface, and chimney area to grate area, and size of boiler to work 
required of it, to reduce the temperature of chimney gases to the 
minimum ; and, if all conditions are favorable, they should not be 
over 4oo°F. 

In making an evaporative test, it is necessary that the duration 
of test should not be less than for ten hours, and to be of any value 
should be made very carefully, and will usually require the services 
of from two to three extra men to assist the regular attendants, and 
these men should have some knowledge of the duties they have to 
perform. It will be necessary to have three accurate platform 
scales, one for weighing the coal and two for weighing the water. 
For the water two good tight barrels or tanks are required, one on 
each scale, and arrangements should be made for filling and weighing 
them alternately. Sometimes the water is drawn from tanks, the 
dimensions of which have been previously taken and the weight of 
water they hold computed, so that all that is necessary during the 
test is to keep a tally on the number of times each tank is filled, 
then the total weight of water can be computed at end of the test. 

The readings of a water meter on feed pipe have also been taken 
for the water ; but as they are not likely to be quite correct, it is 
preferable to use two tanks, each on a separate platform scale, and 
take the actual weight of each as it is filled alternately. 

A reliable thermometer should be placed in a tee in feed pipe, 
near where it enters the boiler, and another in the uptake to chimney, 
and one also in the pipe conveying the feed water into the tanks on 
scales. 

When commencing the test, say at 7 a.in., steam should be up to 
the usual pressure, the ashpit and furnace all cleaned out, and a light 
fire of wood laid on grate ; the tubes all cleaned, and the height of 
water in gauge glass marked by tying a piece of string round it at 
the point where water reaches up to. The attendants should be on 
hand, each having sheets of paper properly ruled off for recording 
the readings of the various gauges, thermometers, etc. This should 
be done every fifteen minutes. 

Several boxes of coal should be weighed out previous to com¬ 
mencing the test, so that the fireman may have a little ahead, and if 
any is left when test is over it can be weighed and deducted from the 
total. 

The utmost care should be taken in weighing the coal and water, 
in taking the readings of the different gauges and thermometers, if 
a correct test is wanted, and to obtain this none of the attendants 
should have more to do than can be done easily. 


88 


engineers’ manual. 


If the plant shuts down at noontime the drafts may be closed, 
fires carefully banked and some person left to look after them, who 
must not allow pressure to exceed the average or safety valves to 
blow off, if possible ; but it is much preferable, if it can be arranged, 
to continue the operation of the plant till the end of the test. 

When the time comes to close the test, the water feed should 
have been so adjusted that the water in gauge glass is just up to the 
string which was tied on glass at starting, and any water remaining 
in weigh tank should be weighed and deducted from the last entry. 

Any coal remaining should also be weighed and deducted from 
the last entry of weighing. 

The fire in furnace should be hauled out and weighed, and its 
weight deducted from the total weight of coal consumed. 

The ashes should also be weighed, and note taken of their 
weight. 

The net weights of both water and coal should then be carefully 
added up, and entered in following manner :— 

Test of.Boiler at Messrs.... 

Day of.. 18 - 

Dimensions. 

No. of flues and diameter. 

Size of fire grate. 

Heating surface. 

Diameter of chimney. 

Height of chimney. 

Duration of test .hours. 

Kind of fuel used. 

Boiler pressure by gauge -lbs. 

Temperature of feed water entering boiler . .. .°F. 

Temperature of feed water entering pump -°F. 

Total quantity of fuel burned.lbs. 

Percentage of moisture in fuel. %. 


Equivalent dry fuel.lbs. 

Total weight of ashes.lbs. 


Equivalent combustible. lbs. 

Total water evaporated.lbs. 

Water evaporated per hour.lbs. 

Water evaported per lb. dry fuel. lbs. 

Water evaporated (per lb. dry fuel) from and at 
212°F.lbs. 

Water evaporated per lb. of combustible from and at 
2 12°F .lbs. 

Horse power developed.H.P. 

To find the % of moisture in fuel, take a fair sample of it and 
weigh it, then let it dry for 24 hours and weigh it again when dry, 
then the difference between the wet and dry weights multiplied by 100 
and divided by the wet weight of the sample will give the percentage 
of moisture. 

To find the water evaporated per hour, divide the total quantity 
of water evaporated by the duration of test in hours. 















engineers’ manual. 


89 


To find the water evaporated per lb. of dry fuel, divide total 
quantity of water evaporated by the total quantity of dry fuel burned. 

To find the equivalent combustible, subtract the weight of ashes 
and clinker from the total weight of fuel burned. 

To find the equivalent dry fuel, multiply the total quantity of fuel 
burned by the % of moisture and divide by 100, then subtract the 
quotient from the total quantity of fuel burned. 

To find the quantity of water evaporated from and at 2i2°F. 
(this is the usual standard), multiply the total heat or heat units in 
1 lb. of steam at average pressure maintained during test (less the 
total heat of 1 lb. of feed water before entering the pump), by the 
quantity of water evaporated per lb. of fuel and divide the product 
by 966, which is the total heat units contained in 1 lb. of steam at 
212°F. This is t called the equivalent evaporation, and is used to 
reduce tests to a common standard for comparison. It is expressed 


thus 


W'= W 


H-f 


966 


-== equivalent evaporation. 


W — lbs. of water evaporated per lb. of coal. 

t° = temperature of feed as supplied (calculated from zero). 

H — total heat of steam in B , T,H, U, at average pressure of test. 
W'= The equivalent evaporation from and at 212 °B. 


To find the H. P. developed, subtract the total heat units of i/lb. 
of feed water before entering the pump or injector, as the case may 
be, from the total heat units in 1 lb. of steam at average pressure of 
test, and multiply the product by the quantity of water evaporated 
per hour, and divide by 1103.4 (which is the heat units necessary to 
raise 1 lb. of water from ioo°F. and evaporate it into steam at 70 lbs.) 
and this quotient divided by 30 will give the H.P., as decided at the 
Centennial Exhibition. 


The following is an example of finding the equivalent evaporation 
from and at 2i2°F. : 

Water evaporated per lb. of fuel = 8 lbs. 

Average temperature of feed water = 4o c F. 

Average pressure by gauge = 6o lbs. 

Total heat of 1 lb. of steam at 60 lbs. = 1175.71 heat units. 

Total heat of 1 lb. of feed water at 4o°F. =8 heat units. 

Then TI 75 - 7 * * j — q. 73 lbs. from and at 2i2°F. 

966 

In making these tests great care must be taken in the details, for 
if any guess work is allowed the test becomes worthless. 

THE INJECTOR. 

Injectors are chiefly used for locomotives, these being seldom 
fitted with feed pumps in modern practice. Injectors will draw water 
from 2' to 12' feet, according to size, but the water supply must be 
continuous and must not be hotter than 135 0 F. for low pressures, 
and 105° F. for the highest pressures. If these temperatures are 
exceeded, so much water is required to condense the steam that the 




engineers’ manual. 


90 

velocity of the steam is too much reduced in driving forward the 
large volume of water. 

Steam is admitted to the injector through a conical nozzle, and 
its admission is regulated by a spindle, the lower end of which fits 
accurately into the nozzle. The water with which the boiler is to be 
fed enters the injector on the opposite side from the steam and 
through a branch a little below the steam pipe branch. 

By admitting steam and water by their respective branches, the 
steam is able to drive the water into the boiler against a pressure 
which is equal to, or it may be greater than its own. This may seem 
paradoxical, but, nevertheless, it is the case, and the explanation is 
as follows: The velocity of an issuing jet of steam is many times 
greater than that of a jet of water issuing under the same pressure, 
and if steam, while issuing from the boiler, be condensed to water, 
but not reduced in velocity to that of the water issuing under the 
same pressure, it is then capable of overcoming the pressure of the 
water in its own boiler. This is exactly what takes place in the 
Gifford’s injector: The steam enters the injector, and passing down 
the conical nozzle is condensed on coming into in contact with the 
feed water, without losing its velocity, further than that due to the 
friction of the passages. The vacuum formed in the injector by the 
condensation of the steam, causes more water to rush into the injector 
and this feed water is carried on by the force of the condensed steam 
jet into the boiler. 

HEATING OF FEEDWATER. 

A due regard for economy in the production and saving of power 
requires that that contained in the heat of exhaust steam be applied 
to some useful purpose, and as a rule is best utilized in raising the 
temperature of the feedwater to the highest point of which it is 
economically capable. To effect this the heater is used ; and when 
in addition to this duty it is possible, by its use, to eliminate most of 
the impurities contained in the water, its great value to an economical 
steam plant will be acknowledged and appreciated. 

That the feedwater heater is a most important feature in a steam 
plant can be very easity proved by the following : Boiler pressure, 
60 lbs. gauge. Feedwater, 40° before and 200° after it goes through 
heater. What is the percentage gained by using the heater ? 


Temperature of steam at 60 lbs.pressure = 307 

Latent heat units in steam at 60 lbs.= 899 

Total heat units. = 1206 


The total heat supplied per lb. of steam is =1206-40, if there were 
no feedwater heater= 1166 heat units ; but feedwater heater increases 

1 ,, ^ . 160 x 100 

the temperature from 40 to 200 or 160 gain in heat -— ^55— = j 

* 3 ' 7 l %- 

By increasing the temperature of the feed from 40 to 200 there 
is a gain of 13.71%. 

To Find the Percentage Gain by Heating Feedwater— 

Rule : Divide 100 times the difference between the final and 









engineers’ manual. 


91 


initial feed temperatures by the total heat units in the steam minus 
the initial temperature of the feed. 

Formula 100 ^ em P« °f feed—Initial temp, of feed ^ 

L Total heat units in steam—Initial temp, offeed J 
Example—Initial temperature of feedwater, 45 0 ; final tempera 
ture, 210 0 ; steam pressure, 100 lbs. gauge. Find % gain. Ans.—14.1% 
The following table shows the per cent, saving by 1 eating the 
feedwater at 60 lbs.: 


Initial 
temp, 
of water 

Final temperature of feedwater. 

120 

140 

160 

180 

200 

250 

300 

35 

7 • 2 5 

8.96 

10.66 

12.09 

14.09 

18.34 

22.60 

40 

6.85 

8-57 

10.28 

12.00 

I 3 - 7 I 

17.99 

22.27 

45 

6-45 

8.17 

9.90 

11.61 

13-34 

17.64 

21.94 

5 ° 

6.05 

7 - 7 1 

9-50 

11.23 

13.00 

17.28 

21.61 

55 

5- 6 4 

7-37 

9.06 

10.85 

13.60 

16.93 

21.27 

60 

5 • 2 3 

6.97 

8.72 

10.46 

12.20 

16.58 

20.92 

65 

4.82 

6.56 

8.32 

10.07 

11.82 

16.20 

20.58 

70 

4.40 

6.15 

7.91 

9.68 

n -43 

15-83 

20.23 

75 

3-98 

5-74 

7 - 5 ° 

9.28 

11.04 

15.46 

19.88 

80 

3-55 

5 - 3 2 

7-°9 

8.87 

10.65 

15.08 

19.52 

85 

3.12 

4.90 

6.63 

8.46 

10.25 

14.70 

J 9 - 1 7 

90 

2.68 

4-47 

6.26 

8.06 

9-85 

14.32 

18.81 

95 

2.24 

4.04 

5-84 

7- 6 5 

9.44 

13-94 

18.44 

100 

1.80 

3.61 

5 - 4 2 

7 - 2 3 

9-03 

13-55 

18.07 

110 

.90 

2 -73 

4-55 

6.38 

8 .20 

12.76 

17.28 

120 

.00 

1.84 

3 - 67 

5 - 5 2 

7-36 

n -95 

16.49 

130 


.92 

2.77 

4.64 

6.99 

11.14 

15.24 

140 


.00 

1.87 

3-75 

5.62 

10.31 

14.99 

156 



•94 

2.83 

4.72 

9.46 

14.18 

160 



.00 

1.91 

3.82 

8-59 

13-37 

170 




.96 

2.89 

7.71 

12.54 

180 




.00 

1.96 

6.81 

11.70 

200 




* 

.00 

4-85 

9-93 


Example- The initial temperature of the feedwater is 85° F., and 
the final temperature 180°. Find the per cent, gained if gauge pres¬ 
sure is 60 lbs. Ans. — 8.46%. 

PUMPS. 

To Find the Capacity of a Pump, per Stroke, in Gallons— 

Rule : Multiply the area of the cylinder by the length of the 
stroke and divide by 277.27. 

Area of cylinder x length of stroke 
Formula, = 

Z) 2 x.7854xZ Z> 2 x L (1) 

277.27 " 352.8 























92 


engineers’ manual. 


To Find the Capacity of a Pump, per Stroke, in Lbs.— 

Rule : Multiply the area of the cylinder by the length of stroke 
and divide by 27.727. 


Formula, 


Area of c ylinder x length of stroke 

27.727 

Z> 2 xZ 
= 35 - 28 


( 2 ) 


To Find the Capacity of a Pump in Gallons, per Minute— 
Rule : Multiply (1) by number of strokes per minute. 

D 2 L x No. of strokes 


Formula, 


35 2 • 8 


( 3 ) 


To Find the Capacity of a Pump in Lbs. per Minute— 

Rule : Multiply (2) by number of strokes per minute. 

^ . D^L x No. of strokes 

formula, -= 

35 - 28 

(3) multiplied by 10. 


( 4 ) 


To Find the Horse-power required to raise water a given 
height— 


Rule : Multiply the volume in cubic feet per minute, by pres¬ 
sure per square foot and divide by 33000, or weight of water in lbs. 
x height of lift divided by 33000. 


Formula, 


Vols. in cub. ft. per minute x press, per sq. ft. 
33000 


Weight of water in lbs. x height of lift 
33000 

Certain allowance should be made for friction, etc., varying from 
15 to 25%. 


Example—What power is required to raise 600 cubic feet of water 
per minute, lifting it 20' and then forcing it to 140' in height. 


Total height of water to be raised =140 + 20=160' 

Total weight of water to be raised =600 x 62.4=37440 lbs. 

160 x 37440 

= 181.5 H.P., and allowing 25% for friction gives 


33 °°° 


us 227 H.P. 


The height of a column of water is equal to pressure per square 
inch •— .433 = pressure per square inch x 2.309. 

Example—What power is required to raise 1000 gallons of water 
per minute, lifting it 20' and forcing it against a pressure of 60 lbs. per 
square inch. Allow 25% for friction. Ans.—60 H.P. 









ENGINEERS’ MANUAL. 93 

To Set the Valves of a Worthington Duplex Pump. 

The steam valve of this pump has no outside lap, consequently, 
while in its central position, it just covers the steam ports leading- to 
opposite ends of the cylinder. To set the piston in the middle of its 
stroke, open the drip cocks and move the piston by prying on the 
crosshead (not on the lever), until it comes into contact with the 
cylinder head ; make a mark on the piston rod at the face of the 
steam end of the stuffing box follower ; move the piston back to con¬ 
tact stroke at opposite end. Make second mark on piston rod half¬ 
way between first mark and the follower. Then if the piston is again 
moved back until second mark coincides with face of same follower, 
it will be exactly at the middle of its stroke. Bear in mind that one 
piston moves valves on opposite side. ( a) When the steam valve is 
moved by a single valve rod nut, as is the case with pumps having 
less than io-inch stroke. Place one piston in the middle of its stroke; 
disconnect link from head of valve rod on opposite side ; then set the 
valve in its ‘ central position ’; place valve nut evenly between jaws 
on back of valve; screw valve rod in or out until eye on valve rod 
head comes in line with eye of valve rod link ; then reconnect. Re¬ 
peat the operation on opposite side and the valves will be properly 
set. ( b ) When the valve rod has more than one lock nut, as is the 
case with pumps having io-inch stroke and over. Place one piston 
in the middle of its stroke and opposite side valve in ‘ central position ’; 
adjust lock nuts, allowing about T % inch ‘ lost motion ’ on each side of 
jaw. Do not disconnect the valve motion. Repeat operation on 
opposite side. By ‘ lost motion * is meant the distance a valve rod 
travels before moving the valve; or, if the steam chest cover is off, 
the amount of ‘ lost motion ’ is shown by the distance the valve can 
be moved back and forth before coming in contact with the valve rod 
nut. To divide the ‘ lost motion ’ equally move valve each way until 
it strikes the nut or nuts, and see if port openings are equal. It is 
advisable that both pistons be placed at the middle of their strokes 
before touching either slide valve. When the stroke of a pump is too 
long, that is, when piston strikes the heads, the ‘lost motion’ should 
be reduced ; contrariwise, when the stroke is too short, increased 
‘ lost motion ’ will tend to lengthen it. 

STRENGTH OF SOLID ROUND SHAFTING. 

The resistance to tension in solid round shafts is directly propor¬ 
tional to the cubes of their diameters, when made of the same material 
and quality. This is evident from the fact that the shaft must offer a 
moment of resistance or shearing moment equal to the twisting moment 

n 

at the instant of rupture. The area to be sheared is = — d? when d 

— diam. of the shaft. The mean arm or leverage at which this re¬ 
sistance acts is equal to half the radius of the shaft for at the centre 
the leverage is = O and at the circumference it is equal to the radius 

7 * d 

of the shaft. The mean arm is therefore ==— = —. 

2 4 

Let .S’ = shearing resistance per square inch of cross section of 
the material, P — force applied at the end of the lever or circumfer- 


engineers’ manual. 


94 


ence of tne pulley ; R = the radius of wheel, a pulley, or length of 


P. P —S ( area of shaft x —) 


=s( 


7 t d\ 

~ d 3 x - 

4 4 




S-^-d 3 is the total shearing moment, when 5* is a constant quan- 
16 

tity for any given material, and 7 t and 16 are also constants. 

• PxP varies as d 3 . 


At the instant of rupture the strength of the shaft just balances or 
is equal to the twisting moment P. P. 

The strength of the shaft varies as d 3 . 

Example—A good wrought iron shaft i" diameter has been found 
to withstand a tongue (P. P) of 8oo foot pounds. What force acting 
at the circumference of a pulley 20" diameter will break a shaft of the 
same material 2" diameter ? 


Let T 1 =P 1 xP 1 = 800 ft. pds. 

T^P.xR^P.x 1 -^- 

12 

P 1 P 1 : P a P*:: d[ : d\ # 

From which we get P X P\ x — -^2-^2 x 

3 

Or P 2 — PlRx X D ‘ 2 ^7680 lbs. 

P^xD\ 

POWER TRANSMITTED BY SHAFTING. 

The amount of power that a shaft will transmit safely is directly 
proportional to the speed at which it is driven ; thus, a shaft that will 
transmit 6 H.P. at 50 revolutions per minute will transmit 12 H.P. 
equally safe at 100 revolutions, and so on. The amount of work per¬ 
formed is obtained by multiplying the force exerted by the space 
through which it is exerted, and therefore if the space varies the 
power transmitted or work done must vary to the same extent. 

Example — If a 3" shaft transmits 20 H.P. safely at 100 revolutions 
per minute, what H.P. may be transmitted by a 4" shaft running at 
80 revolutions per minute? 

Formula, d\.P^ x P^D\R^P x 

when Z) 1 .Z) 2 represents the diameters of 1st and 2nd shafts respec¬ 
tively, P^P^ represents the revolutions per minute of 1st and 2nd 
shafts respectively ; P^P^ represents the H.P. transmitted by 1st and 
2nd shafts respectively. 


From which we get P 2 = P = 37.926 H.P. 

D\Rx 

Example—Find the H.P. that can be transmitted by a good 







engineers’ manual, 


95 


wrought iron shaft 4" diameter, when driven by a wheel 3' diameter 
running at 100 revolutions per minute. The shaft is not to be strained 
above tV of its ultimate strength, i.e. , factor of safety is 10. The 
maximum or rupturing twisting moment that a 1" shaft will withstand 
is 800 ft. pds. Ans.—97.48 H.P. 

The following table shows the power that steel shafting will 
transmit : 


Revs. 

per 

min’te 


50 
60 
70 
80 
90 
100 
110 
120 
130 
140 
150 
160 
170 
180 
190 
200 
22S 
250 
275 
300 
325 
350 
400 
450 
5 °° 


Take 70% of the above powers for wrought-iron shafts. 

SIZE OF PULLEYS. 

Let D — driving pulley 
d — driven pulley 

R=z No. of revs, per minute of driver 
r = No. of revs, per minute of driven 
To Find the Size of the Driving Pulley- 


Diameter of shafts in inches. 


1 i 

2 

2 \ 

3 

3 h 

4 

5 

6 

7 

8 

9 


Horse power they will transmit. 


3 3 

8.0 

15.6 

27 

43 

64 

125 

216 

343 

512 

729 

1000 

4.0 

9.6 

18.8 

32 

5 1 

77 

150 

259 

412 

614 

875 

1200 

4-7 

11.2 

21.9 

38 

60 

89 

i 75 

3°2 

480 

717 

1021 

1400 

5.-4 

12 8 

25.0 

43 

69 

102 

200 

346 

549 

819 

1166 

1600 

6.0 

14.4 

28.1 

49 

77 

115 

225 

389 

617 

922 

1312 

1800 

6.7 

16.0 

3 1 * 2 

54 

86 

128 

250 

432 

686 

1024 

H 58 

2000 

7*4 

17 6 

34-4 

59 

94 

T 4 I 

275 

475 

755 

1126 

1604 

2200 

8.1 

19.2 

37-5 

65 

103 

i 54 

3 °° 

518 

823 

1229 

1750 

2400 

8.7 

20.8 

40.6 

70 

111 

166 

325 

562 

892 

i 33 i 

1895 

2600 

9-4 

22.4 

43-8 

76 

120 

179 

350 

605 

960 

1434 

2041 

2800 

10.1 

24.0 

46.9 

81 

I2 9 

192 

375 

648 

1029 

1536 

2187 

3000 

10.8 

25 6 

50.0 

86 

137 

205 

400 

691 

1097 

1638 

2333 

3200 

11 -5 

27.2 

53 1 

92 

146 

218 

425 

734 

1166 

I 74 I 

2479 

3400 

12.2 

28 8 

56.3 

97 

i 54 

230 

45 ° 

778 

1235 

1843 

2624 

3600 

12.8 

30-4 

59-4 

103 

163 

243 

475 

821 

1303 

1945 

2770 

3800 

i 3-5 

32.0 

62.5 

108 

172 

256 

5 °° 

864 

I 37 2 

2048 

2916 

4000 

15-2 

36 6 

7°-3 

122 

193 

288 

5 6 3 

972 

: 543 

2304 

3280 

4500 

16.9 

40.0 

78 1 

i 35 

214 

320 

625 

1080 

1715 

2560 

3645 

5000 

18 6 

44.0 

859 

149 

236 

352 

688 

1188 

1886 

2816 

4009 

55 oo 

20 3 

48.0 

93-7 

162 

257 

384 

750 

1296 

2058 

3072 

4374 

6000 

21.9 

52 0 

101.6 

176 

279 

416 

813 

1404 

2229 

3328 

4739 

6500 

23.6 36.0 

109 4 

189 

300 

448 

875 

I 5 12 

2401 

3584 

5 io 3 

7000 

27.0 

64-0 

125.0 

216 

343 

512 

1000 

1728 

2744 

4096 

5832 

8000 

30 472 0 

140.6 

243 

386 

57 6 

1125 

: 944 

3087 

4608 

6562 

9000 

33-7 

^80.0 

156.2 

270 

429 

640 

1250 

2160 

3430 

5120 

7290 

10000 










































96 


engineers’ manual. 


Rule : Multiply the diameter of the driven by the number of its 
revolutions, and divide by the revolutions of the driver. 
d.r 

Formula, D= g 

To Find the No, of Revolutions of Driver— 

Rule : Multiply the diameter of the driven by the number of its 
revolutions, and divide by the diameter of the driver. 
d.r 

Formula, R = 


D 


To Find the Size of the Driven Pulley— 

Rule : Multiply the diameter of the driver by its revolutions, and ( 
divide by the revolutions of the driven. 

, , D.R 

Formula, 


d =- 


To Find the Revolutions of the Driven Pulley. 

Rule : Multiply the diameter of the driver by its revolutions, and | 
divide by the diameter of the driven pulley. 

, D.R 

Formula, r --j- 

d si 

To Find the Value of a Train of Gears or Pulleys— 

Multiply the radii, diameters, or number of teeth of all the drivers, 
and divide by the product of all the radii,diameters, or number of 

teeth of the followers. „ , . c . 

Note-The value of a train is the ratio of the number of revolt.- 
tions of the last wheel in a train to the number of revolutions of the 
first wheel in the same train. 

Example-Find the value of the following train : The drivers are 
ACE , having respectively 40, 60, 80 teeth ; the followers are BD , 
having 100, 120, .60 teeth. If F makes 4° revolutions per minute, 
how many revolutions does A make . 4.8, 192. 


Let 


SCREW CUTTING. 

t— Threads per inch to be cut 
y_ “ “ “ on leading screw 

d x d„ = Number of teeth in drivers 
y^y^ — “ “ if “followers. 

The number of threads per inch are inversely proportional to the 
distance between any two consecutive threads. 

t f x x f„ pit ch of leading screw 
~7 r “ d x x d% “ pitch of screw to be cut. 

If the train of wheels is a simple one, we have 


-4r = 4 from which we get 
T d 


t =44 or the number of threads per inch to be 


d 


cut is equal to threads per inch on leading screw multiplied by the 


- 









engineers’ manual. 


97 

number of teeth in follower divided by the number of teeth in the 
driver. 

„ . . t. d 

By transposition we get f =—— or 

To Find the Number of Teeth in the Follower— 

Rule : Divide the product of the number of threads to be cut per 
inch and the number of teeth in the driver by the number of threads 
per inch on the leading screw. 

Example I.—Calculate the number of teeth to be put on leading 
screw in order to cut 12 threads per inch, when the pitch of the lead- 
; ag screw is and the driver on lathe spindle has 40 teeth. Ans.— 
*0 teeth. 

Example II.—Screw to be cut to have 40 threads per inch. Lead¬ 
ing Screw pitch, and using a driver d x of 20 teeth, determine the 
rest of the gears. 

t f x x f ^ 4 0 4x10 80 x 100 

T ~ d x x — 4 1x4 20 x 40 

fi x ft 80+ 100 
d x x d 2 20 x 40 

The wheels required are therefore /' 1 =8o teeth, f 2 = 100, ^=40 

teeth. 

Example I.—Leading screw pitch. Screw to be cut pitch. 
What wheels would you use ? Ans. — 50 driving a 20, or in that ratio. 

Example II.—The set of change wheels belonging to a lathe con¬ 
sists of the following :—20, 25, 30, 35, 40, 45, 50, 60, 70. 80, 90, 100, 
1 ro and 120 teeth. If the pitch of the leading screw is l" devise suit¬ 
able trains to cut the following screws and make a table of your 
results : -4, 4^, 5, 5^, 6, 6£, 7, 8, 9, 10, 12, 14 and 16 threads per inch. 

THE PENDULUIT. 

The time of a single small oscellation is the same time as that of 
a body falling through half the length of the pendulum multiplied by 

P g 

7t. But the space traversed by a body === See formula for fall¬ 

ing bodies. 

2S 

or P — — 

g 

V 2 S 
1 = g 

But the length of the pendulum = 2 s. 


VI 



Therefore, the time T of a single small vibration = Tt — , or 











9 8 


engineers’ manual. 


the time is proportional to the square root of the length of the pendu¬ 
lum while g remains constant. 

To Determine the Force of Gravity — 

VI 

From the Formula T = tc - we get 

g S 

7T 2 / 

g = and if T is taken as i second 
then g — 9.87/ when /is the length of the pendulum 

in feet. 


If a pendulum 39.15" long oscillates once in a second, find the 
value of^-; i.e. the acceleration due to gravity. 

g= 9.87 i 

= 9-. 8 7_ x 39_'5_ =32 . 2feet , 

12 


Let t = 
Wv* 
g r 

h = 


The Pendulum Governor. 
time of one revolution 
centrifugal force 

height of the cone 
radius 



£ 


' Suppose the governor for the time being to be as shown by the 











ENGINEERS’ MANUAL. 


98A 


relitz Tent 
Awning Co., me. 

U7-U9 YESLER WAY 


Awnings, 

Tents, 

Flags, 


Bags 


Mops, 

Tarpaulins, 

Cotton Duck, all 

weights and widths 





98B 


engineers’ manual. 




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Succeeding : 

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engineers’ manual. 


99 


diagram, then by the principle of moments we have 
™L.k-Wr 


h =€l! or_2 = 


The velocity = —or t — 


v* 

VI = r 

g v 
2 Ttr 
1 


g 


2 TCr 


but 


v/T 


r 

=~, and multiply both sides of the equation by 27 T 
-,V~h 2 Ttr 

we get 2it ——— — —— — time of one revolution, which shows that 

the time of one revolution varies directly as the square of the height 
of cone h. 

The equation as given for the pendulum is T = and for 

_ g 

the pendulum governor t — 2 it —which shows that the time of one 

g 

revolution of the governor is equal to one double swing of the 
pendulum. 

Example I.—What length of pendulum, in inches,will oscillate once 
in a second in London. 

t = 7 t 'LL = 39.15" 

g 

Example II.—What is the height of the cone of a governor that 
will make 30 revolutions per minute ? 


t — 27T 




^it 2 h 


g= 


t<i — 


h = 


- JE - =39.1 s" 

4 7T 2 = 7T 2 ^ 5 


which is the same as the pendulum that beats seconds. 

To Find the Height of a Revolving Pendulum which makes a 
given number of revolutions per second— 

Formula, ^ 4- Revolutions per second squared 

4*2 

9 . 78 " 

Revs, per second 2 

Example—What is the height in inches of the cone of a pendulum 
governor making 120 revolutions per minute? Ans—2.445". 

L. of 0. 








TOO 


engineers’ manual. 


ACCELERATION DUE TO GRAVITY. 


Let /—Lime in seconds 

£-=measure of acceleration 
s=space traversed 

w=initial velocity at beginning of interval of time 
7/=final “ “ end “ “ " “ 

v- u— change of velocity in t units 

v ~ ^ = rate of change of velocity =g 

v —u+gt (0 

. u + v 

Average v = - 

2 

Space=average velocity x No. of seconds 



t — u t + 


2 


( 2 ) 

(3) 


If U — O or body starts from rest, the 

Space = - g — - 
2 

In equation (i) v — u+gt 


t= V - - U . and if we substitute this value of t in equa- 

g 

tion (2) we get s =- from which we get 

2g 

v2=U*+2gS (4) 

and if u=o then v 2 =2gs 

v =V 2 & S (5) 

When a body is thrown vertically upward with an initial velocity 
u, to find the space and velocity v at the end of time t ; g in this 
case, is opposite to the motion of the body : 

v — u-gt (6) 

s = ut-?L (7) 

2 

To find the time, /, to reach a given height, s, when thrown with 
a velocity, u, and also the velocity, v, at a given height : 

From equation (7), we get 

t = (g) 

g 

Both values of t are positive. The lesser gives the time required by 
the body to reach the given point, and the greater the time required 
by it to come to rest and fall back to this point. 











engineers’ manual. 


IOI 


.substitute in equation (4) -g for g, and we can find the velocity 
at a given height: 

V 2 = U 2 — 2gS 
V — y/v - 2gS 

when 5 = height. 

Examples — Body thrown vertically downward with a velocity of 
50' per second. Find the velocity at the end of 15 seconds and when 
it has fallen 600'. Ans. — 530' ; 247'. 

A body is thrown upward with a velocity of 160' per second. 
Find the velocity at the end of 3 seconds, and also when it has risen 
400'. Ans. — 64'; o'. 

A body falls from rest for 4 seconds. Find the distance fallen, 
and also the distance fallen in the 4th second. Ans. — 144'; 112'. 

Body projected vertically upward at an initial velocity of 160' pe, 
second. Find distance traversed in 5 seconds, and also the distance 
traversed in the 5th second. Ans.—400'; 16'. 


How to Proceed to Set Up a Stationary Engine. 

Having come to a decision as to the proper position to place the 
engine, drop a plumb line from several points on the line shaft to 
the floor, strike a line on the floor under shaft to correspond with the 
plumb lines and locate where the centre line of engine is to be— 
being sure that it is at right angels to the line already on the floor; 
the best method to adopt to get the two lines at right angels to each 
other, is fully illustrated by the 47th problem of Euclid, “ The sum of 
the squares of two sides of a right angle triangle is equal to the square 
on the hypotenuse” : 

Th 3 square of A added to the square ot B is equal to the square of 
C ’, as illustrated by Fig. 1 ; to apply this, 
measure off on the line under line shaft 12' 
from the point where the centre line of 
engine crosses t and from the same point 
measure off 16'on the centre line of engine, 
the 16' point is to be moved sideways until 
it is exactly 20' from the 12' point on centre 
line under line shaft, then stretch a line from 
the point where the lines cross each other, 
directly over the 16' point, the two lines will then be exactly at 
right angles to each other ; if there is not room to use the distances 
12’, 16' and 2o r , use 6', 8' and 10', but it is best to use the first, as with 
this any slight deviation would only be one-half as much as it would 
be with the last. To get the position of outer bearing and have it 
square with centre line of engine, proceed as before ; or, measure 
from the centre line under line shaft at two different points to get the 
crank shaft of engine parallel with this line. The excavation for 
foundation and pier of outer bearing can now be made—it should be 
2' or 3' wider and longer than the intended foundation When the 
excavation is finished, the templet for the anchor bolts may be set; 
care should be taken that the centre line on templet is directly under 
the line representing centre line of engine. If the bolts are to be 






102 


engineers’ manual. 


built in, they can be hung in templet, but the best practice is to have 
hand-holes in bottom of foundation, in which case drop plumb lines 
through holes in templet and lay off the hand-holes 12 wide, and 
when the foundation is built up 12" from bottom, lay oak plank 3 or 
4" thick, of sufficient length to reach over the two hand-holes; when this 
is in position, again drop the plumb line and bore holes in the plank 
i" larger than the anchor bolts ; the holes in foundation from oak 
plank to top of rubble stone or brick work can be left by using a taper 
stick or piece of pipe or square wooden box ; the size °* holes will 
depend on the size of bolts required, but should be 1 or even more 
larger than the bolts. Should the taper sticks be used, care should 
be taken to turn them occasionally to prevent the cement adhering to 
them ; in any case, the foundation should be built with Portland 
cement and good clean sharp sand: if bricks are used, they should 
be dipped in water before being laid. The depth and weight of 
foundation will depend on the size and weight of engine ; if the bot¬ 
tom on which the foundation is to be built is very soft, there should be 
18" or 24" of concrete (broken stone and cement), put in on which the 
foundation should rest; the base of foundation should be wider and 
longer than the top, that is, with a batter on the side of walls about 
1^" to the foot in height. When the top stones are laid, care should 
be taken that they are perfectly level on top and the cement used 
under these stones slightly stronger than the other part of founda¬ 
tion. As soon as they are placed in position, the engine can be placed 
and lined up ; to do which, in most cases it is best to remove the 
piston and pass a line through cylinder, taking great care that the 
line is in the centre of cylinder at back end and the centre of stuffing 
box at front end also passes over the centre of crank pin ; by turning 
the crank shaft and adjusting outer bearing so that the centre of 
crank pin is directly over the line at both back and forward centres, 
the shaft will be at right angles with centre line of engine. The ad¬ 
justment of engine and outer bearing, to get them level, should be 
done with iron wedges and when in proper position and the fly-wheel 
on shaft, sulphur may be run under engine frame cylinder and 
stand for outer bearing, the anchor bolts tightened, piston put in 
cylinder and connections made. There are several more details that 
might be mentioned, but they would not apply to every case, but as 
they occur, the engineer in charge of erecting should be prepared to 
meet them. 

To Set Up a Wheelock Engine. 

After the engine is placed on foundation, level the frame length¬ 
wise by placing a level on lower guide and level it the other way by 
dropping a plumb line over the four high points left on the frame in 
front of guides for this purpose, place one thickness of paper between 
the line and the two points on upper guide, this will enable the man 
doing the job to see how close the line is to the high point on lower 
guide,* and when right the engine will be level both ways. - Place the 
level in the main bearing, the levelling should be done with wedges 
under cylinder and crank end of frame. The stand and outer bearing 
can now be placed and the crank shaft put in position ; the stand 
adjusted with wedges until shaft is level. To line up the engine, make 
line fast to centre of crosshead pin and pass it forward half length of 


ENGINEERS MANUAL. 


I03 


crank pin from face of crank, turn shaft to back and forward centre 
to see that it is at right angles with centre line of cylinder ; fly-wheel 
can then be placed in position on shaft, keyed up and allowed to stand 
a few hours while making steam connections, then try the shaft and 
frame again to see that they are level and that the foundation has not 
settled, and if all right run sulphur under cylinder, crank end of frame 
and stand for outer bearing, tighten the foundation bolts, put on con¬ 
nection rod, finish steam and exhaust connections then engine is ready 
to start. Should the valves need adjusting, look on the face of arms 
on cut-off valves for small marks for this purpose. 

When the valves are at rest a fine 
plumb line should drop directly over the 
points A and B , Fig. 2 ; should this not 
be right, it can be made so by loosening 
the set screw which holds the stud on 
which the dash pot rests, the end of this 
stud is eccentric, so by turning it the 
dash pot can be lowered or raised as 
the case may require ; when right, be 
sure to tighten set screw. In order to 
get the lead point, place the crank on 
the forward or back centre as the case 
may be, being careful that it is correct 
and the hook holding the valve in posi¬ 
tion, a fine plumb line should fall directly 
over the points C and D , Fig. 2. If 
after indicating the engine the diagrams 
show that the load is not equally divided 
between both ends of cylinder, the rod between the two trips should 
be lengthened or shortened as the case might require. In starting 
the engine for regular work, it would be well to allow an extra quan¬ 
tity of cylinder oil to pass through lubricator for at least a week after, 
by this means the piston and inside of cylinder will become polished 
and be less liable to cut. 

RIVETTED JOINTS. 




Let P— pitch of Rivets in ins. 
d— diam. 

/= thickness of Plate “ 

Ts = Tensile strength of 
Plate per sq. in. 

55 = Shearing strength of 
Plate per sq. in. 

7j=No. of rows of Rivets, 






















engineers’ manual. 


104 

The strength of the plate between rivet centres is = Px lx Ts. If 
at the ends of the line, P, two holes are drilled d' in diameter, the plate 
is diminished by two halves of one diameter or one diameter, i.e ., 
P-d=s, which represents the available section, and the strength 
becomes ( P—d ) lx Ts, obviously less than Pxtx Ts. 

The ratio of the strength of the plate; after the holes are drilled, 

to that of the original plate will be represented by 1 ^^ — — 5 and 

the percentage of the strength of the plate will be -—-. Fron* 

this follows the 


Rule for finding the percentage of strength of plate after the 
holes are drilled : Divide 100 times the difference between the pitch 
and the diameter of the rivet by the pitch. 


Example—Pitch of rivets, 2^" ; diam. of rivet=f". Ans.—70%. 

The shearing strength of a rivet — Area of rivet x Shearing 
strength ; and if there are, n, rows of rivets, then 

Area of rivet x x n— Shearing strength of n rivets. 

But P. t. Ts = original strength of plate. 

Area of rivet xSsxn 

- rt . rp - = Ratio of strength of rivets to that 

P' t ls of original plate. 


or 


Area of rivet xSsxnx 100 
P. lx Ts 


— % strength of seam. 


And if we assume Tl=Ss, then 


Area of rivet x n x 100 

~¥ 7 t 


= % strength of seam to that of 
original plate. 


Example—Thickness of plate, §" ; rivets, i|"diam. ; pitch, 3I". 
If seam is double rivetted, find the percentage of strength of rivets 
to that of original plate. Ans.—71%. 

The most economical form of a joint is one in which the plate and 
rivet strengths are just equal, and the best way to arrive at this is to 
equate the formula for the rivet section to that of the plate section, 
thus : 

Area of Rivet X No. of Rows of Rivets _ P- d 

. 

or (P- d) Pt = Area of rivets xnx P. 

from which we get (P- d) l = Area of Rivets x n 


Area x n 

or P = - ~ t -- + d 


(0 


To Find the Pitch of Rivets if all the other data is given— 
Multiply the area of the rivet by the number of rows of rivets, 
then divide by the thickness of the plate and add the diameter of the 
rivet to the quotient. 










engineers’ manual. 




If thickness of plate is required, then 
Area of Rivet x n 
t== P-d 

To Find the Area of the Rivet — 

(P-d) t 

- -—— = area 

n 

or if area of rivets is given to find the number of rows 
(P~d)t 


fa 


(3) 


(4) 


As it is more practical to take the diameter instead of the area of 
the rivets the formulae would have to be changed, as follows : 


Formula (i) becoming P = 

( 2 ) 


d* x .7854 x n 


+ d 


f _ d 2 X . 7854 x n 
P-d 


(3) 


(4) 


d 


-V 


(p-d)t- 


(5) 

( 6 ) 

(7) 


• 7854 x n 
(P-d)t _ (P-d) tx 1.273 
d 2 x . 7854 d 2 


( 8 ) 


Examples—Find the pitch of the rivets so as to secure an equal 
percentage of strength in rivets and plate of a double rivetted seam 
plate £", rivets i|" diameter. Ans. — 3.775" 

Pitch 3.775", plate f". seam double rivetted. Find area and 
diameter of rivet. 

(P-d) t 

Take formula (3) when area = - ^ ~ we get .994 sq. inches, 

and from (7) we get 

Diameter of rivet= J (P ~ d '> T=J (3-77S Z l ■ 12 5 ) jlW 

\ . 781:4 X « \ .78a X 2 \3.I4I 


7854 X 


975 

3- H l6 


=y/ 1.2652 = 1.125" or 1 

To prove that this joint is correct, all that is necessary is to find 
the percentage of strength of the plates after the holes are drilled to 
that of the original plate, and also the shearing strength of the rivets 
to that of the original plate ; or, 


P-d 


=70% in the above example ; 


Z) 2 x.78t;4x« Area of rivet xn _ n/ 

or - ‘—^1 - ---- =70% 

P.t P.t 

Professor Unwin gives the diameter of the rivet as being = 
1.2^, and which is considered a very good proportion ; and from 
this we can easily calculate the pitch and diameter of the rivet, if the 
thickness of the plate is known. 

Example — Plate, f" thick. Find P and d in a single rivetted lap 
joint. 






















io6 


ENGINEERS MANUAL. 


The diameter of the rivet is found to be i .044", and by formula 
(1) we get the pitch to be 2.445". 

The percentage of strength of rivets in a single rivetted lap joint 
compared with that of the original plate = 

Area of rivet x n x ioo_ .994 x 100 Q/ 

-- n — j. --- q —54/0 

Pxt 2 .445 x £ 

and the strength of the plate after holes are drilled to that of the 
original plate is ^^ -=54% 

The strength Of a single rivetted lap joint is only 54% of 
the original plate. 

Example in Double Rivetted Joint — Plate, £" thick. Find pitch 
and diameter of rivets ; also, percentage of strength of joint to that 
of the original plate. 

The diam. of the rivets is the same as in the single rivets, 
viz. = 1.2 y /1 = ij" nearly; and P will be found to be 3.795", in 
practice 3I". 

If the joint is properly designed, the strength of the rivet section 
will be equal to that of the plate section, and in this instance it will 
be found that the strength of the rivet section is .7 of the original 
plate. From this, we find that the strength of a double rivetted 
lap joint is 70% of the original plate. 

Tripple Rivetted Lap Joint — 

Example— Plate £" thick. Find pitch, diameter of rivet, and also 
percentage of strength of joint. The diameter is the same as in the 
single and double rivetted joints, viz., 1.044", an d by formula (1) we 
g-et P—§.or, By calculating the strength of the joint as above indi¬ 
cated we get it to be close on to 80%. Therefore the strength of a 
triple rivetted lap joint is 80% of the original plate. 

The following is a summary of the above : 

The strength of plate between rivet centres =P. t. Ts. 

The percentage of strength of plate after holes are drilled— 

The shearing strength of a rivet is = Area of rivet x .Ss. The per¬ 
centage of strength of rivet section of n Rows to that of original plate 

• Area of rivet xSsxnx ioo_Area of rivet xnx 100 
p. t. Ts. ~p~r 

if shearing and tensile strength are equal. 

The pitch is = Area of rivet x« + n , 

Single rivetted lap joint is .54 of the solid plate. 

Double “ “ “ is .70 “ “ “ 

Triple “ “ “ is .80 “ “ ‘‘ 








engineers’ manual. 


107 

STRENGTH OF BOILER SHELLS. 

The question of the form of boiler was unimportant while very 
low pressures were used, but as soon as the higher pressures from 
40 lbs. per square inch and upwards were adopted it became neces¬ 
sary to devote eonsiderable attention to the form of shell that would 
best withstand internal pressure. 

The sphere is the strongest form for any boiler, but owing to 
many reasons for not adopting this in practice the cylindrical boiler 
is universally used as being the nearest approach to the sphere. 

The force tending to rupture a boiler is = the pressure per square 
inch x by the diameter in inches, and is arrived at in the following 
manner : 

Let P = bursting pressure per square inch 
t — thickness of plate in inches 
D = diameter in inches 
Ts = tensile strength of the material. 



In the diagram take the 
small surface A B which lies 
at the angle @ with the line 
CD. 

By the resolution of forces 
the bursting pressure P on 
surface A B can be resolved 
into two components, viz,: 
Pv and Ph , which are the 
vertical and horizontal com¬ 
ponents. 

Pv — P. cos 0 
The vertical pressure on 
A B=A B. P. cos ©, and cos 
©. AB — ab. 

From this we see that the 
vertical pressure on the surface A B = P. ab, and as the sum of all 
the vertical components of P will be = P x diameter. 

If the boiler has a length l we have a pressure tending to cause 
rupture = P. D. L. and we have resisting this pressure 2 t. 1 . T s. and 
therefore at the point of rupture the resistance of the material = 


pressure tending to cause rupture. 


From this we get P. D. 1 . = 2 t. 1 . 

PD = 2t.Ts 
p_2t. Ts 
D~ 


.*. t = 


P.D 
2 Ts 


Ts 


(1) 


We see from this that the pressure P required to cause longitud¬ 
inal rupture is equal to twice the thickness of the plate multiplied by 
the tensile strength of the material divided by the diameter of the 
boiler in inches. The boiler may also be ruptured transversely due to 
the pressure on the ends. In this case the force tending to cause 






io8 


engineers’ manual. 


rupture = pressure x area of boiler (cross section) i.e., Px.D-. 7854 
and the resistance of the plate = area x Ts at the point of rupture. 

P.D‘>^-=TtD.t.Ts 

^pJ -tf.Ts (2) 

D 

From (1) and (2) we see that the pressure causing rupture longi¬ 
tudinally is one-half that causing rupture transversely, or 

P _ 2t.Ts 
~P D _ 1 
4 1. Ts 2 

and for this reason the longitudinal seams are always made stronger 
than the circumferential seams. 

In boilers having internal flues the pressure required to rupture a 
boiler transversely is greater than twice that for the longitudinal rup¬ 
ture for the area exposed to pressure is less than the whole cross sec¬ 
tion of the boiler by the area of the flues. 

Example— A boiler 28' long, 7' diameter, has two 30" internal 
flues running the whole length of the boiler, the thickness of the plate 
is The longitudinal seams are double rivetted and the transverse 
seams single rivetted. Find the bursting pressure along the longi¬ 
tudinal and transverse sections. Tensile strength of plate 50.000 lbs. 
square inch. Ans. —417 lbs. 1534 lbs. 

STAYS. 

To Find the Strain on Each Stay — 

Rule : Find area of surface supported, and multiply it by 
pressure carried, and divide it by the number of stays, and the 
quotient will be the strain on each stay. 

To Find the Proper Diameter of Stay— 

Rule : Strain on one stay divided by 5000 (which is the strain 
allowed per sq. in. of stay), then divide by .7854, and extract the 
square root, will give the diameter required. 

To Find the Proper Pitch of Stays — 

Rule : Strain on one stay, divided by steam pressure to be 
carried, and extract the square root, will give the proper pitch of 
stays. 

To Find the Total Amount of Stays in Square Inches Necessary— 

Rule : Area of sheet to be stayed, multiplied by pressure to be 
carried, divided by 5000. 

Knowing pitch of stays and steam pressure, to Find Strain on One 
Stay— 

Rule : Pitch squared and multiplied by pressure = strain on one 
stay. 

To Calculate the Area of a Diagonal Stay — 

Rule : First find the area of a direct stay sufficient to support 




engineers’ manual. 109 

the surface to be stayed, then the area of direct stay, multiplied by 
the length of diagonal stay, and divide the product by the length of 
a line drawn at right angles from the surface to be stayed to the 
point where diagonal stay is secured. Thus— 



WATER AT DIFFERENT TEMPERATURES. 

The component parts of water by weight and by measure are— 

By Weight. By Measure. 

Oxygen . 88.9 1 

Hydrogen . 11 . 1 2 

One cubic inch of water at its maximum density, 39 .i°F., weighs 
252.6937 grains, and one cubic foot weighs 62.425 lbs.; it is 828.5 
times heavier than atmospheric air. 

The four notable temperatures are — 1. Freezing point, 32 0 F. ; 2. 
Maximum density, 39 .i°F. ; 3. Standard of specific gravity, 62° F.; 
4. Boiling point at sea level, 212 0 F. 

Below 39. i° it decreases in density very slowly at first, and very 
rapidly as it nears the point of congelation . A cubuc foot of ice weighs 

57.25 lbs. and expands .089=—-— of its bulk. 

r Imperial. U. S. 

One cubic foot of waterj §ub£ 7 nch'es. 7J gallons. 

[62.4 lbs. 

One gallon of water... j ^ 7^74 cubic ins. 23^ cubic inches. 

It has the greatest solvent power of any of the liquids ; for 
common salt this is constant for all temperatures. For other 
matter, such as carbonate of lime, magnesium and the different 
sulphates, its solvent power increases as the temperature increases. 
It decreases in weight as the temperature increases — 

At 32 0 F., weight per cubic foot 62.418 lbs. 

“ 39- T ° F > “ “ 62.425 “ 

“ 62° F., “ “ 62.355 “ 

“ 212° F., “ “ 59 - 7 6 ° “ 

Water is practically incompressible, and its capacity for absorb¬ 
ing heat is greater than any other liquid or solid. 












IIO 


engineers’ manual. 


The following- table gives the heat units per lb. and the weight of 
a cubic foot of water at temperatures from 32 to 212 F.: 


Temp. 

Fah. 

Heat 
units 
per lb. 

Weight 

per 

cubic ft. 

Temp. 

Fah. 

Heat 
units, 
per lb. 

Weight 

per 

cubic ft. 

Temp- 

Fah. 

Heat 
units 
per lbs. 

V/eight 

per 

cubic ft. 

32 0 

o.co 

62.42 

114 

82.02 

61.83 

154 

122.34 

61.10 

35 

2.02 

62.42 

115 

83 02 

61 82 

155 

12334 

61 08 

40 

8.06 

62.42 

Il6 

84.03 

61.80 

156 

124 35 

61.06 

45 

13.08 

62.42 

117 

85 04 

61 78 

157 

125 36 

61.04 

5 ° 

18 10 

62.41 

I l8 

86.05 

61 77 

158 

126.37 

61.02 

52 

20 II 

62 40 

1 1 9 

87 06 

61.75 

159 

127.38 

61 00 

54 

22.11 

62.40 

120 

88 06 ~ 

61.74 

160 

128 38 

60 98 

56 

24 II 

62 39 

I 21 

89 07 

61 72 

l6l 

129 39 

60 96 

58 

26.12 

62.38 

122 

90.08 

61 70 

162 

130.40 

60 94 

60 

28 12 

62.37 

323 

91.09 

61.68 

163 

X 3 X - 3 X 

60 92 

62 

30.12 

62.36 

124 

92 . IO 

61 67 

164 

132 42 

60.90 

64 

32.12 

62.35 

125 

93 10 

61 65 

165 

x 33 32 

60.87 

66 

34 12 

62 34 

126 

94 11 

61 63 

166 

x 34-43 

60.85 

68 

36 12 

62 33 

I27 

95 12 

61 61 

167 

x 35 34 

60.83 

7 ° 

38 n 

62 31 

128 

96.13 

61 60 

168 

x 3 6 -35 

60.81 

72 

40.II 

62.30 

129 

97 14 

61.58 

169 

x 37 46 

60 79 

74 

42.II 

62.28 

130 

98 14 

61.56 

170 

138 46 

60 77 

76 

44.11 

62.27 

X 3 X 

99 15 

61.54 

17 c 

x 39-47 

60 75 

78 

46.10 

62.25 

132 

100.16 

61 52 

172 

140 38 

60.73 

80 

48.09 

62 23 

133 

IOI I7 

61.51 

x 73 

141 49 

60.70 

82 

50 08 

62.21 

134 

102 18 

61.49 

*74 

142 50 

60.68 

84 

52 07 

62.19 

135 

103 18 

61.47 

x 75 

x 43 5 o 

60 66 

86 

54.06 

62 17 

136 

104.ig 

61 45 

176 

x 44 5 X 

60 64 

88 

56.05 

62 15 

137 

105 20 

61 43 

177 

145.62 

60.62 

9 ° 

58.04 

62.13 

x 3 8 

106 21 

61 41 

178 

x 46.53 

60.59 

Q2 

60.03 

62 n 

139 

I07 22 

61.39 

x 79 

x 47 54 

60.57 

94 

62 02 

62.09 

140 

I08 22 

61 37 

180 

148 54 

60.55 

96 

64.01 

62 07 

141 

log 23 

61.36 

181 

x 49 55 

60.53 

98 

66.01 

62.05 

142 

IIO.24 

61 34 

182 

150.56 

60.50 

100 

68.01 

62.02 

143 

III 25 

6x 32 

i8 3 

X 5 X •57 

60.48 

102 

70.00 

62.00 

144 

112 26 

61 30 

184 

152-58 

60 46 

104 

72.00 

61.97 

i 45 

II3.26 

61 28 

185 

x 53 58 

60 42 

106 

74 00 

61 95 

146 

II4.27 

61 26 

186 

x 54 59 

60 41 

108 

76 00 

61.92 

i 47 

115.28 

61.24 

187 

x 55 - 6 o 

60.39 

no 

78.00 

61.89 

148 

1 l 6 29 

61.22 

188 

156 61 

60.37 

112 

80.00 

61.86 

149 

117 30 

61 20 

189 

157.62 

60.34 

Xx 3 

81.01 

61.84 

1 5 ° 

H8.3O 

61.18 

190 

158.62 

60 32 




i 5 r 

ng 31 

61 16 

191 

1 59 -63 

60.29 




x 5 2 

120.32 

61.14 

192 

160.63 

60 27 




153 

121.33 

61 12 

J 93 

161.64 

60.25 



1 




194 

162.65 

60 22 







x 95 

163.66 

60.20 







196 

164.66 

60.17 







197 

165.67 

60.15 







1 198 

166.68 

60.12 







199 

167.69 

60.10 







200 

168.70 

60.07 







201 

169 70 

60.05 







202 

170.71 

60.02 







203 , 

171.72 

60.00 







204 

x 72 - 7.3 

59-97 







205 

x 73-74 

59-95 







206 

x 74-74 

59-92 







20 7 

x 75-75 

59 89 







208 

176.76 

59.87 







209 

177.77 

59.84 







210 

178.78 

59.82 







211 

179.78 

59-74 







212 

180.79 

59-76 






























engineers’ manual. 


Ill 


Mechanical Refrigeration and Ice Making. 

In order to arrive at a clear understanding as to what processes 
of nature are applied in refrigerating and ice-making machines, atten¬ 
tion is called to certain facts well-known by everybody. If, especially 
on a warm and dry summer day, the hand is moistened with water 
and then exposed to a current of air, a decided sensation of cooling 
will be noticed. The dryer the air and stronger the current, the 
more will be the cooling effect. The explanation of this phenomenon 
is that by the rapid circulation of air over the wet hand the water is 
evaporated, and that for this evaporation it needs heat, which is to a 
great extent taken from the warm hand and therefore produces the 
cooling effect upon the skin. If instead of water the hand is mois¬ 
tened with alcohol the cooling effect will be still greater because 
alcohol is so much more volatile than water, and if sulphuric ether is 
used for this experiment the effect will be still more marked. The 
heat which is required to transform a liquid into a gas is called the 
“ latent" heat. What becomes of the heat which is constantly sup¬ 
plied to boiling water? The answer to this is that it is used to change 
the liquid condition of the water into that of gas or vapor, and the 
heat supplied in this manner not showing any increase on the ther¬ 
mometer must be contained in the vapor. It appears again if the 
vapors are condensed into a liquid, and this proves that while the 
water was in the vaporous state the heat contained in it to maintain 
it in that condition has become “latent.” 

This physical law of nature is made use of in nearly all the 
machines which have for their object the reduction of temperatures. 
If the substances which we are in the habit of using for this purpose 
were to cost nothing, we could simply let the evaporated agents, such 
as ammonia, carbonic acid, sulphuric ether, or dioxide of sulphur, 
escape into the atmosphere. But since the substances are rather 
costly, it becomes a matter of necessity for the purpose of economical 
refrigeration to maintain them in the apparatus which is used for the 
purpose of cooling or ice-making, and all the cumbersome machinery 
which is used for the production of cold has really no other object than 
to restore the refrigerating agent by alternately liquifying it and re¬ 
evaporating it. During the process of evaporation it takes up the 
latent heat from the surrounding bodies and thus cools them, and in 
the process of condensation this heat is again given off to the cooling 
water that is showered over the condensers and by it carried away 
into the thermal ocean. To give an idea how much a pound of ice, 
or its equivalent in cooling, would cost by using ammonia if it was not 
retained in the apparatus ; it should be stated that the latent heat of 
water being 142 and the latent heat of ammonia about 560, it would 
take about one pound of liquid ammonia evaporated to produce the 
cooling effect of the melting of about 4 pounds of ice (3^ lbs. in reality 
counting certain losses). The price of ammonia at 30 cents a pound 
would make the equivalent of 1 lb. of ice cost 7 \ cents or $150 per ton, 
and this simple calculation proves that it is absolutely necessary to 
use the refrigerating agent over and over again, and to provide an 
apparatus which, in the waste of this agent is as economical as pos¬ 
sible, or, in other words, have an apparatus which is as nearly per¬ 
fectly tight as possible. 


112 


engineers’ manual 


Of all the agents to-day used for purposes of cold production 
ammonia has so far maintained its superiority. Under ordinary tem¬ 
peratures ammonia is a gas, but it can be liquified by compression 
and cooling. After liquifaction and being exposed in an open vessel 
to the ordinary pressure of the air, it will boil, at a temperature of 
of 27°F. below zero, which makes it eminently fit for the production 
of low temperature. It is a non-inflammable substance of absolute 
stability and will, if contained in an apparatus built entirely of iron or 
steel, retain its chemical composition for all time. Carbonic acid is 
also, to a very limited extent, used for the production of cold, but it 
has not all the excellent qualities which ammonia possesses, especially 
on account of the enormous pressure necessary to liquify it, and on 
account of the quality it possesses of not being liquifiable at all above 
a temperature of about 90° F., while ammonia can be liquified at a 
much higher temperature than this. Besides, where condenser 
pressures in machines using ammonia generally range from 150 to 180 
lbs. per square inch, the condenser pressures of the carbonic acid 
machines lie between 800 and 900 lbs. per square inch ; and while the 
evaporating pressure of the ammonia is generally in the neighbor¬ 
hood of about 25 lbs. per square inch, that of carbonic acid machines 
is as high as 200 to 250 lbs. per square inch, so that the piping 
system—cocks, all joints and the compressor—have to be a great 
deal stronger than is necessary with the use of ammonia. 

There are two different systems for the handling of ammonia, 
one is called the Compression System and the other the Absorption 
System. The former uses the refrigerating agent in its purest state 
and entirely free from all watery admixtures, and is, therefore, 
anhydrous. The Absorption System uses the solution of ammonia in 
water, and is denominated the Absorption System on account of the 
final operation by which the evaporated ammonia is regained. In 
the Absorption machine the ammonia is driven out of its solution in 
water by being heated by means of a steam coil ; and the ammonia 
thus driven off and carrying along with it a certain percentage of 
water is condensed in an apparatus called the “ Condenser,” which 
is kept cool by water being supplied to the outside surface of the 
pipes in which the ammonia is forced from the “Still.” The liquid 
ammonia, after being collected in a receiver or storage tank, is 
passed through a cock into pipe coils, called the “ Refrigerator,” 
in which a lower pressure than that of the condenser is maintained. 
The latter is accomplished by passing the ammonia gas generated 
in the refrigerator into another vessel, called the “Absorber.” In 
this vessel the gas is again brought into contact with the water from 
which it had been driven out before by heat, the water before it 
enters the absorber being cooled down "to as low a temperature as 
the cooling water of Nature will permit. The combination of the 
ammonia gas and this water again liberates heat, and therefore the 
absorber has to be kept cool by running water the same as the 
condenser. The strong - solution is now by a little feed pump pumped 
back into the boiler, or still, and the cycle of operations is started 
anew. 

In the compression system the process is simpler. By the suck- 
ing action of the compressor pump the ammonia gas is drawn away 


engineers’ manual. 


Ir 3 

from the refrigerator coils, and is compressed on the return stroke of 
the piston into the condenser—condenser and refrigerator coils being 
practically the same as those of the Absorption System. From the 
condenser to liquified ammonia is likewise passed through a small 
regulating or expansion cock into the refrigerator, and the cycle of 
operations here commenced again. It thus appears that there is one 
less operation during the process through which the ammonia 
passes, and that is the absorption of the gas. 

It is very easily understood now, as the evaporation of the 
ammonia produces the cold in the refrigerator pipes, that these pipes 
can be utilized in any manner desired for the cooling of other bodies 
They may be put up in a room in which the air is thus directly cooled, 
or they may be put into a tank with salt brine, which is non-con- 
gealable, except at a very low temperature, and by their cold sur¬ 
faces cool the brine, which in return may be circulated through pipes 
in a room, and thereby produce a lowering of temperature. The 
former is called the “ Direct Expansion System,” and the latter is 
called the “ Brine Circulating System.” 


Properties of Ammonia. 


Temp.deg. 

Fah. 

Gauge press. 
Lbs. per sq. in. 

Heat of 
vaporization. 

Volume of vapor 
per lb. cub. ft. 

Weight of a 
cub. ft. of vapor 
in lbs. 

40 

0 

579-67 

24-37 

.0410 

35 

0 

576.69 

21.29 

.0467 

30 

0 

573-69 

18.66 

•0535 

25 

1.47 

570.68 

16.41 

.0609 

20 

3-75 

567.67 

14.48 

0690 

15 

6.29 

564.64 

12.81 

.0779 

10 

9 °7 

561.61 

11.36 

.0878 

5 

12.87 

558.56 

10.12 

.0988 

0 

i 5- 6 7 

555 50 

9.04 

. 1109 

5 

19.47 

552-43 

8.06 

. 1241 

10 

23-85 

549-35 

7-23 

.1384 

15 

28.23 

546.26 

6.49 

• r 54 o 

20 

33-25 

543-15 

5-84 

. 1712 

25 

38-73 

540-03 

5.26 

. 1901 

30 

44 7 1 

53692 

4-75 

.2105 

35 

5 1 23 

533 78 

4-31 

.2320 

40 

58 30 

530-63 

3 - 9 1 

.2583 

45 

65.96 

527 47 

3-56 

.2809 

50 

74-25 

524-30 

3-25 

• 3 io 9 

55 

82.93 

521.12 

2.96 

•3379 

60 

92.90 

5 I 7-93 

2.70 

•3704 

65 

103-33 

5 * 4-73 

2.48 

•4034 

70 

114 5 1 

5 11 52 

2.27 

•4405 

75 

126.55 

508.29 

2.08 

. 4808 

80 

i 39 4 1 

504.66 

1.91 

.5262 


If ice is wanted, the cold brine which is produced by the refriger- 





















engineers’ manual. 


”4 


ating coils may be used to have immersed in it galvanized iron cans 
containing pure water; and if the brine is kept at a temperature of, 
say, 14 0 or 15 0 below the freezing-point of water, it is evident that, 
after more or less time, the water in the can will finally be frozen. 

After it is entirely frozen, the can is lifted out and the ice melted 
out by applying tepid water to the outside of the can. Thus a block 
of ice is formed of the exact shape of the can. 

HORSE=POWER. 

The unit of power universally adopted by mechanical engineers 
in this country is that which was proposed and used by Watt, viz., the 
horse-power. What is technically spoken of among engineers as a 
horse-power is the rate of doing work corresponding to 33,000 foot 
pounds per minute, and the power of engines is always calculated on 
this basis. 

The horse-power exerted by an engine is = total mean pressure 
on the piston in pounds multiplied by the distance in feet travelled by 
the piston in one minute, divided by 33,00a 

Let H. /^Indicated horse-power 

iV=Number of strokes per minute=Revolutions x 2 
£=Length of stroke in feet 

-^^Area of the cylinder in square inches=Z? 2 x .7854 
/ 3 =Mean effective pressure in pounds per square inch. 

The mean pressure on the piston is = P. A and the distance 
travelled in one minute by the piston is = L.N 


P. L. A . N 

33 °°° 


(1) 


.’. the horse-power of the engine = 

Ard by transfer— 

To find the mean effective pressure when all other data are given : 


Formula, P=z— 'f > ' 33 °o° 

L.A.N 

To find the length of stroke in feet : 

Formula, L — ^’^' 33 °°° • 

P.A.N 

To find the area of the cylinder : 

Formula, A = LL 33 °°° 

’ P.L.N 

To find the diameter of the cylinder : 

Formula, D— v ^^‘^ >x 33 °o-— 

P.L.N x .7854 

To find the number of revolutions : 

Formula, No. of revolutions 33 °°° 


(2) 


( 3 ) 


( 4 ) 


s/ LL.Px 42000 (5) 

P.L.N 


P.A.L 


( 6 ) 


Examples What is the horse-power of an engine running at 300 
revolutions per minute 18" stroke, diameter of cylinder 10", and the 
mean effective pressure 50 lbs. ? Ans.—21.42 horse-power. 

What is the mean effective pressure in an engine of 120 horse- 











engineers’ manual. 


!I 5 


power, running at 150 revolutions per minute, 15" diameter of cylin¬ 
der, and stroke 30 '. Ans.—30 lbs. 

The mean pressure as indicated by a diagram taken from an 
engine running at 100 revolutions per minute is 45 lbs. per square inch, 
the area of the cylinder is 100 square inches. Find the length of the 
stroke if the engine is developing 87^ horse-power. Ans.— 3' 2 

What is the diameter of the cylinder of an engine which is run¬ 
ning at 300 revolutions per minute, 20" stroke, mean pressure 50 lbs., 
and indicates 225 horse-power. Ans. - 13.75 horse-power. 

To Find the Horse-Power of a Compound, Triple, or Quadruple 
Expansion Engine, calculate by the rule given above the horse-power 
of each cylinder separately and then add the results. 

Note : Treat each cylinder as if it were a separate engine. 

Example—A compound engine having cylinder areas in the ratio 
of 1 14, is running at 120 revolutions per minute. The stroke is 36", 
diameter of high pressure cylinder is 20", and the mean effective 
pressure on the low pressure cylinder is 15 lbs. per square inch. What 
is the horse-power of the engine ? Ans.—834.86 horse-power. 

DUTY OF AN ENGINE. 

The duty of an engine is the number of footpounds of work done 
by the consumption of too lbs. coal. In 1891 a committee of the 
A. S. M. E. recommended a new unit, viz.: footpounds of work per 
million heat units furnished at the boiler. This is equal to the old 
unit when the coal imparts 10,000 H.U. to the water in the boiler or 
to the evaporation of 10.35 lbs. from and at 212 0 per pound of coal. 

Taking the old unit, the duty of a pumping engine that will do 
100,000,000 footpounds for every 100 lbs. coal burned is said to be 100 
million. 

Example I. — An engine requires 3 lbs. coal per 1 H.P. hour. 
What is the duty ? Ans. — 66 millions. 

Example II.—The area of a pump plunger is 100 square inches, 
double stroke 4', number of double strokes 9600, coal burned 800 lbs. 
Gauge pressure on main pipe shows 50 lbs. and the height of this 
eauee is 24. i' above the water in the well. Find the duty. Ans.— 
28,800,000. 

BRAKE HORSE-POWER. 

It is often advisable to know the actual power given out by an 
engine independent of the power absorbed in friction, etc., in driving 
the engine itself. In order to do this it is necessary to either apply 
an absorption or a transmission dynamometer to the flywheel, or to a 
pulley keyed on the crank or first shaft. The power so obtained is 
termed the brake horse-power (B. H. P.) One of the simplest and 
most easily applied absorption dynamometers is that known as the 
Prony Brake. 

The formula for finding the H. P. is as follows : 

2 TtrnP 

H.P. - = -0001904 xrxnxP 

33000 . 

when r — radius of pulley or horizontal distance from centre of shaft 
to centre of weight. 

n = number of revolutions per minute 
P— pull or weight in pounds. 



n6 


engineers’ manual. 


It is important to observe that neither the diameter of the pulle 
nor the pressure of the friction blocks enter into the formula. 
Example-Cylinder 7" diameter, stroke 10", pressure 55 lbs. r = 2.5 
n = 624, P = 96. Ans.— H.P. = 28.52. 

HORSE=POWER TRANSMITTED BY BELTS. 

The ultimate strength of ordinary bark-tanned single leather 
belting varies from 3000 to 5000 lbs. per square inch of cross section. 
For convenience sake the tenacity is stated generally in lbs. per inch of 
width. For single belts the breaking stresses are from 750 to 1250 
lbs. per inch of width. For double belts the breaking stresses are 
from 1500 to 2500. Allowing for joints this is reduced to about one- 
third of the strength of the solid leather and allowing a factor of safety 
of 5, we get a safe working stress of r V of the breaking stresses of 
the solid leather. 

For single belts the working tension would be from 50 to 80 lbs. 
“ double “ “ “ “ “ “ “ 100 “ 160 “ 

In practice , however, 50 lbs. for single and 80 lbs. for double belts 
per inch of width are considered about the maximum. Under these 
conditions leather belts will run for many years. 

Horse-Power that Leather Belts will Transmit Per Inch 
in Width at Various Speeds. 


By A. G. Brown, M.E. 


Velocity 
of Belt 
in feet 
per 

minute. 

Best Oak Tanned Belts. 

Velocity 
of Belt 
in feet 
per 

minute. 

Best Oak Tanned Belts. 

Single. 

Light 

Double. 

Heavy 

Double. 

Single. 

Light ! 
Double. 

Heavy 

Double. 

100 

• *5 

.21 

•27 

2100 

3-i8 

4-45 

5-73 

200 

■30 

.42 

•55 

2200 

3-33 

4.67 

6.00 

300 

•45 

.64 

.82 

2300 

3-49 

4 • 88 

6.27 

400 

.61 

•85 

1.09 

2400 

3-64 

509 

6-55 

5 °° 

.76 

1.06 

1.36 

2500 

3-79 

5-30 

6.82 

600 

• 9 1 

1.27 

1 .64 

2600 

3-94 

5-52 

7.09 

700 

1.06 

1.49 

1.91 

2700 

4.09 

5-7 3 

7 36 

800 

i .21 

1.70 

2.18 

2800 

4.24 

5-94 

7.64 

900 

1.36 

1.91 

2-45 

2900 

4-39 

6.15 

7 ' 9 ‘ 

1000 

15 1 

2.12 

2-73 

3000 

4-50 

6.36 

8.18 

1100 

1.67 

2-33 

3.00 

3 IO ° 

4.60 

6.58 

8-45 

1200 

1.82 

2-55 

3 -27 

3200 

4.69 

6.79 

8.70 

1300 

1.97 

2.76 

3-55 

33 oo 

4-77 

7.00 

8.86 

1400 

2.12 

2.97 

3.82 

3400 

4.84 

7.21 

8.96 

1500 

2.27 

3.18 

4.09 

35 o° 

4.90 

7-3i 

9.06 

1600 

2.42 

3-39 

4-36 

3600 

4-95 

7.40 

9.16 

1700 

2.58 

3-6i 

4.64 

3700 

4-99 

7.48 

9.24 

1800 

2-73 

3.82 

4.91 

3800 

503 

7-54 

9.29 

1900 

2.88 

403 

5 -18 

39 °° 

5 05 

7.60 

9-34 

2000 

3-03 

4.24 

5-45 

4000 

5.08 

7.64 

9-37 




























engineers’ manual. 


TI 7 


Effect of Centrifugal Force on Belts. 

When belts or ropes are run at high speeds the tensions in the 
two parts of the belts or ropes between the pulleys are greater than 
that calculated from the horse-power transmitted. This increase of 
tension is due to the centrifugal force set up in those parts of the belt 
which are in contact with the pulleys. 

The centrifugal force has the effect of diminishing the normal 
pressure between the belt and the pulley and therefore of diminishing 
the resistance to slipping. 

ENERGY OF A ROTATING FLYWHEEL. 


The energy possessed by a moving body is called kinetic energy, 
and its amount for a falling body is obtained by finding the height 
through which it must fall to acquire the velocity of its motion. If 
this height be obtained, the. work is equal to the height in feet x 
weight in pounds, or, if w = weight in lbs. and h = height in feet, the 
work don e = Tvh foot pounds. 

v 2 

In falling bodies, h = - ; g— the acceleration due to gravity 32' 

|2 


per second, and t;-=velocity in feet per second. 


V‘ 

if h— - then 

2 g 


7vh — 


2 g 


-, which is equivalent to saying, if a body of w lbs. moves 


with a velocity of v feet per second, the energy or accumulated 


work = 


22 - 


Example—The rim of a flywheel weighs 2 tons, and the mean 
velocity 30' per second. How many foot pounds of work are stored 
up in it? Ans.—56250 foot pounds. 

Rule : Multiply the weight in pounds by the velocity in feet per 
second squared, and divide by 64. 

Another example—Flywheel weighs 5 tons ; mean radius of 
rotation 5' ; 100 revs, per minute. Owing to the load being sud¬ 
denly diminished, the speed increases to no revs, per minute. 
What reserve power is stored up in the flywheel to overcome any 
sudden increase of the load ? Ans.—89375 foot pounds. 


CENTRIFUGAL STRESS IN FLYWHEELS. 

There is no flywheel made that will not burst if it were only run 
fast enough. The centrifugal or centre flying force in the arms is 
directly proportional to the velocity squared, so that by doubling the 
velocity the centrifugal force or stress is quadrupled. 

To Find the Centrifugal Force— 

Rule I.: Multiply the weight in pounds by the velocity in feet 
per second squared, and divide by the radius multiplied by 32. 

Wv 2 

Formula, -- = centrifugal force. 




1/8 


ENGINEERS’ MANUAL. 




Rule II.: Multiply the product of the weight and velocity squared 
by .03125, and divide by the radius. 

Formula, .03125 -= centrifugal force. 

Rule III.: Multiply the product of the weight and radius by the 
revolutions per minute squared, and then by .000342. 

Formula, — JVxr. revolutions per minute 2 x .000342 = 
centrifugal force. 

Example—A flywheel 12' mean diameter weighs 6 tons and runs 
at 70 revolutions per minute. Find the centrifugal force. Ans.—121000. 





BURSTING STRESS IN FLYWHEELS. 

With centrifugal force the pressure is acting radially in all direc¬ 
tions and is analogous to the pressure in a boiler. If we wish to 
determine the effective force tearing apart the rim of the wheel or the 
shell of a boiler we must find the resultant of this force acting in one 
direction and upon one-half of the rim. Take the last example under 
the heading of Centrifugal Stress, where the centrifugal force = 
Wv 2 

-= 121000; this stress is supported at two points in opposite 

sides of the rim, and in order to break the wheel two pieces or both 
sections must burst, therefore the stress on each side = 60500 lbs. 
Wv 2 


= stress on each side trying to pull the two halves asunder. 

The weight of a cubic inch of cast iron weighs fully a quarter of a 
lb. or .26 lb. The number of cubic inches of iron in the rim will be 
= 46000 , and the mean circumference of the rim is 144 x it = 452". 
4.6oOO 

= 102 st I uare inches in the rim. If we now divide the stress 

on each side by the area multiply by Tt we get the bursting stress per 
square inch ; 

60500 

-—= 188 lbs. per sq. inch. 

102 x Tt r n 

From this, we get the following formula : 

wv 2 


2 Area ^= Burstin S stress P er s fl- inch * 

This formula can be simplified as follows : 

Let D = diameter of wheel in feet. 
r — radius in feet. 

« = No. of revs, per minute. 
v = velocity in feel* per minute. 

Stress per sq. inch =(rx n ) 2 x .001065 3 

“ “ (approx.)=(r x n ) 2 x .001 

“ “ =(Z) x n ) 2 x .0002664 

“ “ =v 2 x . 00027 

Example — A flywheel 12' diameter, weighing 12000 lbs., runs at 






ENGINEERS’ MANUAL 


118 a 


GEO. H. ISMON, P. C. S. Agent, San Francisco. 

O. D. COLVIN, Sales Agent, Seattle, Wash. 

kricanMM. 

Manufacturers. 

Electrical Wire 

AND 

WIRE ROPE 


NAILS, BARBED WIRE, PLAIN WIRE, BALE 
TIES, ETC. 


Office 108 W. Washington Street, Seattle, 

Warehouse Cor. First Ave. South and Connecticut St. 



118 b 


engineers’ manual 



.;.,‘„;..%.:..:.a.:.A*Nvvv*/vv% m ! , ^v , >< , ^*W , w ‘’”*’ vvvvvvvvvVV ’' 


108 RAILROAD AVE., SEATTLE. 

Builders of Marine Engines 


Compound and Triple Expansion 

As Designed by J. T. Heffernan, 

Mechanical Engineer 






ENGINEERS MANUAL. 


ng 

70 revs, per minute. Find the bursting- stress per square inch. Sup¬ 
pose wheel had six arms, find the maximum number of revolutions 
the wheel could be run at without breaking, neglecting in the cal¬ 
culation the binding strength of the rim. Each arm has a breaking 
stress of 120,000 lbs. Ans.—168 revs. 

BURSTING STRESS OF FLYWHEEL. 

In practice 100' per second is about the maximum which cor¬ 
responds to a stress of about 970 lbs. per square inch. Good cast 
iron has a tensile strength of about 20000 ; therefore at this velocity, 
100' per second, there is a large factor of safety, but, as the centri¬ 
fugal forces increase as v 2 , we find that cast iron will burst at 454' 
per second, as follows : 

Stress per square inch=(velocity in feet per minute) 2 x .00027 
“ “ =(velocity in feet per second) 2 x .0972 

.'. 20000 —\v' per second) 2 x .0972 

•v/20000 

v= — —=454 feet per second. 

ZEUNER’S DIAGRAM. 

Given the position of crank at point of cut-off, the amount of lead, 
travel of valve, to determine the angular advance and amount of out¬ 
side lap. 



Draw two lines at right angles cutting each other at O. With 
centre O and radius = half the travel of the valve describe the circle 
A GCEB R. Draw <9C=position of crank at cut-off. With A as 
centre and radius = lead of the valve, describe part of a circle and 
then draw CD touching this circle. Through O draw O G at right 
angles to CD. Bisect CD G at the point G then draw GOR, On 







120 


engineers’ manual. 


O G describe circle GM 0 F and on O R describe the circle O PQ R N. 
From centre O radius OH describe the circle HFML. This dia¬ 
gram is now completed and we can readily see the distance the valve 
has moved from its central position for any position of the crank, and 
also the opening of the port to steam at that point. Suppose crank 
to be at G O A and having moved in the direction as shown by the 
arrow, the distance O G is the amount the valve has moved from its 
central position and GH represents the opening of port to steam at 
that particular point. The shaded portion GFHM of the diagram 
shows the opening of the port to steam. By the diagram we see that 
when crank is in position OD the valve is just beginning to opening 
and when it reaches the dead centre A B the port is open an amount 
— KL — lead of valve. When crank reaches OF, release, the valve 
has passed its middle position and is distant from it on the other side 
an amount OP — inside lap; therefore at that point the exhaust 
opens and continues to open. When the crank has reached B the 
valve has moved from its centre a distance = OQ, and since OS = 
the inside lap the port is open to exhaust an amount = Q S. When 
crank arrives at R the port is full open to exhaust and when it arrives 
at O T the valve is closed and compression begins. 

Given the travel, the lap, and the angle of advance, to find the 
point of cut-off, the amount of lead, etc. 

Draw the circle AG CE B R to represent the travel of the valve, 
and the circle HFML to represent the outside lap and draw the 
angle G 0 C = angle of advance. Bisect O G and describe the valve 
circle G M O F as before ; we now see that when the crank is in the 
position O A the valve is open an amount equal to KL, therefore KL 
is the lead of the valve. Through the point F where the lap circle 
intersects the valve circle draw 0 C, then O C represents the position 
of the crank at cut-off. 

Given the travel, the lap and the lead, to find the angle of advance. 
As usual draw the circle to represent the valve travel. Lay off O L 
= the lap and K L — lead of the valve. From K erect K G perpen¬ 
dicular to A B ; then the angle GO C is the angle of advance. 

Examples—Travel of valve 8f", outside lap 2^", inside lap angle 
of advance 35 0 . Find the points of admission, cut-off release, com¬ 
pression and the amount of lead. 

Travel of valve 5", outside or steam lap f", angle of advance 22%°. 
Find graphically the position of the crank at admission and cut-off. 

VALVE SETTING (Slide Valve). 

In setting the valves of an engine, it is of primary importance 
that the points of opening of the valves be known and trammed at 
convenient points on valve rod. To do this, the necessary tools are 
a piece of very thin tin, a piece of steel rod 7" long, sharp-pointed 
and hardened at each end, one end being bent square 1 

To Tram the Valve. — Remove the steam chest cover and place 
the tin in the head port, bringing the valve against it ; then from a 
fixed centre point on end of steam chest (not the gland) tram a line 
on valve rod, marking same with a fine centre punch, repeating the 


engineers’ manual. 


121 


operation on the other steam port. After satisfying- yourself that 
steam chest and ports are clear, put on cover and connect eccentric 
rod. Proceed to set valve by placing- engine on one dead centre ; 
then from fixed point on steam chest use tram to mark valve rod, 
turn engine to other dead centre and again mark rod ; then compare 
marks, and adjust by lengthening or shortening rod or turning 
eccentric, as required. 

TO PLACE AN ENGINE ON THE DEAD CENTRE. 

To place an engine on its dead centre, bring the crosshead to 
within about half an inch of the end of its travel. Take a pair of 
dividers, and from a point on the guides strike an arc of a circle on 
the crosshead, and, with the engine in the same position, tram from 
a point on the floor to the rim of the wheel ; then move the engine in 
the direction it is to run until the crosshead has passed the end of its 
travel and returned to a point where the dividers will coincide with 
the mark already made on the crosshead. Make another tram mark 
on the rim of the flywheel, and midway between these two marks make 
a centre punch mark for a dead centre mark, bring the flywheel to a 
point that the point of the tram will just enter the dead centre mark, 
and the engine is on its exact centre at that end; repeat the operation 
on the other end 

In all cases, move the engine in the direction it is to run, and, if 
moved past the dead centre mark, it must be backed up far enough 
to take up the lost motion before reaching the mark again. 

VALVE SETTING. 

Corliss Engine .—Remove the back covers from valve cylinders; 
the lines marked on valve and cylinder ends are lines of opening. 

On the back hub of wrist plate will be found a centre line, a line 
will also be found on the hub of stand which supports wrist plate— 
when these two lines meet, the wrist plate will then be in its central 
position. On either side of centre line of wrist plate stand will be found 
another line, and when centre line of wrist plate coincides with either 
of these lines, it will be in its extreme position. Place the wrist plate 
in its central position and by the means of adjusting nuts make the 
lengths of the valve connections such that each steam valve may have 
the necessary lap £ to §—depending on size of engine, and that the 
exhaust valves may be just opening or without lap. Then adjust 
length of eccentric rod, that wrist plate may vibrate equally. Place 
the crank on any dead centre and turn the eccentric on the shaft in 
the direction engine is to run enough to show an opening or lead of 
say -fa to fa of an inch, then tighten the set screw in eccentric and 
place crank on the other centre and note if lead is the same, if not, 
adjust as required. 

To adjust the disengaging gear, let the governor remain in its 
lowest position, move wrist plate to extreme of travel and hold in this 
position, adjusting cam rods as required. 

To prove the correctness of the cut-off adjustment, raise the 
governor to ifes working plane, blocking it there ; then with eccentric 
connected to wrist plate turn, engine slowly in its running direction, 


122 


ENGINEERS’ MANUAL. 


and measure on the slide the distance the crosshead has moved from 
its extreme position at either end, if cut-off is equalized the distance 


should be the same. . , , 

In all cases it is desirable that an indicator be used to more ac¬ 
curately adjust the setting of £he valves so that the engine may be in 
the best possible condition for economical work. 

Brown Engine.— This is a four-valve engine, the valves being of 
the gridiron type. The most important point is to know the laps 
and openings of the several valves. The steam valves are generally 
marked on the stirrup block, flush with the top of its guide the 
exhaust openings and laps being marked on the end of exhaust rod. 

There being so many points in this valve gear tba.t can b e 
changed, the marks should in all cases be verified. This is usually 
done by getting the point of opening through the peep holes in back 
of steam and exhaust chests, and adjust to original marks by 
lengthening or shortening valve rods, as required. Having marked 
or verified the position of all valves, proceed to set by placing the 
engine on the head end centre, see that gears are secured in position 
to turn engine in desired direction. Engage clutch, then turn head 
eccentric in the direction side shaft is to run until the lower mark on 
stirrup block is above the guide, say of an inch ; secure the 
eccentric. Turn the engine back one-fifth of its stroke, and turn 
exhaust cam ahead until the outside mark on exhaust rod coincides 
with the tram or gauge, with the inner mark approaching same. 
Secure the cam, place the engine on crank end centre, and repeat 
the operation. Then block the governor in highest position, turn 
hand wheel and see if the steam valves trip when the lowest mark on 
stirrup block shows, say, of an inch above guide, setting cut-off 
shaft or levers until it does. Then place governor in lowest position, 
turning hand wheel to see if valves will unhook. Finally, set 
governor in working plane, turning engine to see if cut-off is 
equalized. 


THE SAFETY VALVE. 


Rules for Area of Safety Valves. 


Philadelphia Ordinances— Every boiler when fired separately, 
and every set or series of boilers when placed over one fire, shall 
have attached thereto, without the interposition of any other valve, 
two or more safety valves, the aggregate area of which shall have 
such relations to the area of the grate and the pressure within the 
boiler as is expressed in the following schedule : 

Schedule .—Least aggregate area of safety valve (being the least 
sectional area for the discharge of steam) to be placed upon all 
stationary boilers with natural or chimney draft: Area of combined 
. , 22 . c grate surface in sq. feet 

safety valves =-— — - : - r -— r - ~rs~ZT 

Press, in lbs. per sq. in. above atmosp. +8.62. 

Rule of U. S. Supervision Inspectors of Steam Vessels: 


Lever safety valves to be attached to marine boilers shall have 
an area of not less than 1 sq. in. to 2 sq. ft. of grate surface in the 
boiler, and the seats of all such safety valves shall have an angle of 
inclination of 45 0 to the centre line of their axes. 




engineers’ manual. 


123 


Spring loaded safety valves shall be required to have an area of 
not less than 1 sq. inch to 3 sq. feet of grate surface of the boiler, 
except as hereinafter otherwise provided for water tube or coil, and 
sectional boilers ; and each spring loaded valve shall be supplied 
with a lever that will raise the valve from its seat a distance of not 
less than that equal to one-eighth the diameter of the valve-opening, 
and the seats of all such safety valves shall have an angle of inclina¬ 
tion to the centre line of their axes of 45 0 . All spring loaded safety 
valves for water tube or coil and sectional boilers required to carry 
pressures exceeding 175 lbs. per square inch shall be required to 
have an area of not less than one square inch to six square feet of 
grate surface of the boiler. Nothing herein shall be construed so as 
to prohibit the use of two safety valves on one water tube or coil and 
sectional boiler, provided the combined area of such valves is equal 
to that required by rule for such vaive. 

The Canadian Steamboat Act provides that every safety valve 
must have a lift equal to one-fourth its diameter at least. The open¬ 
ings to and from the valve must each have an area not less than the 
area of the valve, and the area of the safety valve must be equal to 
one-half inch for every square foot of grate surface of the boiler. 

The following are rules for the calculation of the weight, length 
of lever, etc., for safety valves : 

Let W = weight of ball at end of lever in lbs. 
w = “ “ lever in lbs. 

w 1 = “ “ valve and spindle in lbs. 

L — distance from fulcum to centre of W in inches 
/ — “ “ “ i( “ “ valve 

Q = “ “ “ gravity of lever in inches 

A — area of valve in square inches 
P — pressure in lbs. per square inch. 

By the principle of moments we get 

P.A x /= Wx L + w x G + tv x x / (1) 

p WL + w G + w t l 

AJ 

By transposing the equation we get 


TJ7 ._ P.A l-w G-w 1 l 

( 2 ) 

l 

T P A l-w G - w x l 

(3) 

W 



Examples—Find the weight to be placed at the end of a lever 
20" long weighing 15 lbs., the area of the valve being 8 square inches. 
Weight of valve and spindle 6" acting at a distance of 2" from fulcum. 
Steam pressure 60 lbs. Ans.—39-9 lbs. 





124 


engineers’ manual. 


If the weight of the lever and the weight of the valve and spindle 
are omitted from the calculation, the formula becomes 


P= 


Or W- 


WL 
. A l 
P.A l 


And 

“ W 


OF INTEREST TO ENGINEERS. 

By organized and persistent effort engineers can advance their 
interests. 

How many there are who would derive incalculable benefits 
could they be induced to practise what is implied by the above 
heading, but who, by ignoring this, miss many great opportunities, 
and wander or drag along in the same old rut. One may well be 
surprised to find what progress can be made by the end of a year 
by steadfast application for the space of one-half hour each day. 
How much time is lost by aimlessly dreaming and living without any 
solid beneficial subject for thought! A great author has said, 
“ Knowledge is power therefore, if we lack knowledge, we cannot 
properly embrace opportunities presenting themselves, then, by all 
means, let us use organized and persistent effort to secure knowledge. 
We cannot all expect to go to college ; and, bearing this in mind, we 
must remember that the most college can do for us is to put us on 
the road leading to knowledge ; so, those of us who have been 
unable to get a college education should do the best we can to 
advance ourselves by organized and persistent effort. Remember, 
we cannot know it all, as it takes everybody to know everything, and 
very little of anything is yet known. Steer clear of him who claims 
to know it all, for, if you do not, association with such a man will 
have a tendency to disgust you with your fellowmen, and more in 
particular by the way in which he will expose his own ignorance. 

A few things that an engineer should do. —Give his work his 
undivided attention. Give his employer the benefit of his experience. 
Give his attention to the best publication, so that he will advance by 
the experience of others. Give his influence and experience for the 
benefit of his brother engineers, and do all in his power to advance 
the standing of an engineer by being sober, industrious, and never 
forgetting that it is his first duty to look after the best interests of 
his employer. 

What engineers should at all times desire. —Clean engine-room, 
c.ean feedwater, clean coal, clean fire, clean oil. Steady employ¬ 
ment, steady steam pressure, steady feed and steady and regular 
lubrication. Silent action of his engine, steam distribution perfect, 
short watches. High economy, high steam pressure, high expansion 
of steam, and last, but not least, high wages. 




ENGINEERS’ MANUAL 


125 


A. A. GRIFFING IRON CO. 

of 

JERSEY CITY, N, J, 

Inquiry Solicited 
Comparison Invited 

A 

We are pleased to corres^ 
pond with parties using 
steam in any form, 

A 

Simplicity, Durability' 
and Satisfaction are the 
causes of the popularity 
of our Steam Specialtie 

"BUNDY” 

SEPARATORS 

FEED WATER HEATERS 

RETURN AND SEPARATING TRAPS 

EXHAUST HEADS 

LOW WATER ALARMS 

RADIATORS 

STEAM AND HOT WATER HEATING BOILERS 

F. L. PARKER 

General Sales Agent 

417 New York Block SEATTLE 

Phone Red 3956 





126 


engineers’ manual. 


vn\\\\\\nv\\\\\\v\w'W^ wwwwx%x > 

I 

* Light Beat Power 

I ..Electricity.. 


THE PERFECT LIGHT. No Matches, 
Smoke or Odor; Costs no more than the 
Inconvenient Illuminants. 

THE MOST CONVENIENT POWER. 
Easily Handled and Inexpensive. Motors 
of all Sizes, and for all Purposes. 

Wiring, Electrical Supplies, Dynamos, 
Motors, Marine Work, Repairs. 

A complete stock of Combination and Electric 
Fixtures in designs and finishes that harmonize 
with any surroundings. 

PORTABLE LAMPS, GLOBES AND SHADES. 


the Seattle Electric Co. 

907 Tirst Jfvenue. 

I 5 . 



/ 


engineers' manual. 



















































128 


engineers’ manual. 


ELECTRICITY. 


Ohms Law —The strength of a current varies directly as the 
ElecriS motive-force and inversely as the resistance or the intens.ty 
of the current is equal the E.M.F. d.vided by the R. 

E 


C = %i R— Y"< E — CR. 




The unit of resistance is called the Ohm and is equal to io C. . . 
(centimeter, gram, seconds,) units. It is the resistance of a colum 
of pure mercury i square millimetres in section and 106.21 centimetres 
Wr at S Theunit of current is called the ampere, and is io~ 
C G I units. It is that current which will deposit 4.025 grams of 
silver'per hour or decompose .oo 55 944 grams of water per hour. 
The unit of E.M.F. is called the volt and is equal to io 8 C.G.S. units, 
and is the E.M.F. necessary to send a current of 1 ampere through a 
resistance of 1 Ohm. 


iisiauue 1 r 

Resistance.— The resistance of conductors of identical material 
varies inversely as their section end directly as their length ; or, the 
length of one wire multiplied by the diameter squared of the other is 
iqu g al to the square of its own diameter multiplied by the length of 

the other. 


Formula 


2 2 

l\d-i — l*d x 


R , = W 

R. hd\ 


or aV tr; 

when R x d x l x = resistance, diameter and length of one wire 

4 = resistance, diameter and length of other wire. 

From the above we get 


d 2 = 


_ R x h d i 


d, = 


l x d 
^E x l*d\ 


h d\ 


Example—Find the diameter of a copper wire 480' long that has 
twice the resistance of another copperwire 120' long and measuring 
.25 of an inch in diameter. Ans.-.35". 

The total resistance of a wire varies directly with the specifiic 
resistance of the substance of which the wire is made. 

Let S x and 5‘ 2 =specific resistance of wires (1) and (2) 
or K x and K 2 conductances of the “ (1) and (2) 

Then from (1) we get 

R x _ l x d\ x S x _ l x d\ K % 


*2 l 2 d*xS 2 hd\K x 


Or R, 


R x h d \ S * 


l x d\ S x 


R x / 2 d\ K x 














engineers’ manual. 


129 


... Uorthwest... 
Tixture Company 

313 FIRST AVE. SOUTH, 
SEATTLE. 


i 


Electric Supplies and 
Construction 

MANTLES, TILES AND GRATES. 


Gas, electric and Combination 
Tixtures 

TELEPHONE PRIVATE EXCHANGE 6 







130 


engineers’ manual. 


Example—The resistance of a mile of pure copper wire .134" 
diam. is 3.03 Ohms. Calculate the resistance of half a mile of German 
silver wire .0335" diameter. The specific resistance of copper is 1642 
and that of German silver 2'n7o. Ans.—312.52 Ohms. 


Specific Resistance in C GS units at 0 ° C. 


Silver annealed, 1521 CGS units. 

“ hard drawn 1652 “ 

Copper annealed 1615 “ 

“ hard drawn 1642 
Gold “ “ 2154 ‘ 

Zinc 5690 “ 

Platinum anneal’d 9158 i( 

The resistance of a wire .oc 
is 10.4 Ohms, and from this data c 
ances of all other wires. 


Iron annealed 9827 CGS units. 

Nickel 12600 “ 

Tin 13360 “ “ 1 

Lead 19847 “ 

German Silver 21170 “ 

Platinoid 34000 “ 

Mercury 96146 

1" diameter and 1' long- at 6o°F. 
in be readily calculated the resist- 


Example I.—What is the resistance of 1000'of wire .1" diameter, 
knowing that 1 mil. foot (mW' diam - x lon g) has 10 -4 Ohms resist¬ 
ance? 

By transposing formula (1) we get 

R\ 1% d\ = R% l t d\ or 


= 


R 2 l x d\ 10.4 x 1000 x i- 


/, d\ 


I X IOO' 


= 1.04 Ohms. 


Example II.—What length of wire .05"diameter would be required 
so that there would be a resistance of 9 Ohms? Ans.—2163. 


RESISTANCE OF DERIVED CIRCUITS. 


The joint resistance of several circuits in multiple is 


1 

1 1 1 where r 1 r 2 r 3 + = the resistance of each branch. 

r x r 2 r 3 


If there are only two wires in multiple the joint resistance is 



r i r 2 product of theresistances 
r 1 +r 2 ~ sum of the resistances 


Example—Three wires in derived circuit have a joint resistance 
of 6 Ohms. What resistance must be inserted in multiple so that the 
joint resistance will be reduced to 3 Ohms. 

1 


1 1=3 Ohms .*. x = 6 Ohms. 

6 + x 


Example I.—What must be the R of the shunt used with a gal- 
vanometre whose?? is 4500, so that the R of the shunted galvanometre 
shall be 450 Ohms. Ans.^—500 Ohms. 










engineers’ manual. 


131 


Washington Electric Supply 
and mantel Co. 

Incorporated. 

A. R. PINKNEY, Gen. Mgr. 



WHOLESALE AND RETAIL 

Electrical Supplies 

jRnd machinery 


Gas, Electric and Combination Fixtures 

cMantels, Tiles and Grates 

CELEBRATED HALL INCANDESCENT LAMPS. 
DYNAMOS AND MOTORS. 


716 THIRD AYE. TELEPHONE MAIN 745. 




132 


engineers’ manual. 


DIVISION OF CURRENT. 

The relative strength of current in the different branches o? es 
divided circuit is directly proportional to their conductivities, or in 
the inverse proportion to the resistances. 

Example I. — Three wires, 5, 8 and 12 Ohms, are joined in multiple, 
and a current of 49 amperes is sent through the circuit. How much 
will flow through each wire ? 

The joint conductivity of several wires, r lf r 2 , r 3 , in multiple, 
1 1 1 

is = — + — + — + 

r i 7 'A 

F+T+A=nV From this, we see that the current divides, 
as it were, into 49 parts, 

24 of which flow through the wire of 5 Ohms resistance. 

u u u g “ “ 

IO “ “ “ 12 “ “ 

Example II. — A current of 39 amperes is sent through a cirouit 
of 3 wires in multiple having 8, 12 and 16 Ohms respectively. What 
current will flow through the 16-Ohm wire? Ans.—9 amperes. 

ELECTRICAL UNITS OF MEASUREMENT. 


The centimetre = unit of length =. 3937". 

The gram = unit of mass =15.432 grains. 

1 


The second =unit of time part of a mean solar day. 

The sq. cent. = unit of area =.15501 sq. inch. 

The cub. cent. = unit of volume = .061027 cubic inch. 


The unit of velocity is the velocity of a body which moves through 
unit distance in unit time, or the velocity of 1 centimetre per second. 

Momentum is the quantity of motion in a body, and is measured 
by mass x velocity. 


Acceleration is the rate of change of velocity. The unit of 
acceleration is that acceleration which imparts unit changes of 
velocity to a body in unit time, or an acceleration of 1 centimetre per 
second—per second. 

The acceleration due to gravity is considerably greater than 
this = 32.2 feet per second, or 981 centimetres .'. ^=981 centimetres. 

Force is that which produces motion or change of motion in a 
body. The unit of force is that force which, when acting for one 
second on a mass of 1 gram, gives to it a velocity of 1 centimetre 
per second It is called the dyne. 

The force with which the earth attracts any mass is usually 
called the weight of that mass, and the force with which a body 
gravitates, i.e ., its weight (in dynes) is found by multiplying its mass 
(in grams) by the value of g. 

Work is the product of a force and the distance through which it 
acts. The unit of work is the work done in overcoming unit force 
through unit distance, i.e., in pushing a body through a distance of 
1 centimetre against a force of 1 dyne. It is called the Erg. 



engineers’ manual. 


133 



E. W. HOUGHTON 

ARCHITECT 

414-417 Collins Bldg. 

9 


Seattle, Wash. 


134 


engineers’ manual. 


The force with which gravity pulls a mass of i gram is 981 
dynes ; therefore, to lift a mass of 1 gram through a distance of 1 
centimetre is = 981 ergs of work or g ergs. 

Power is the rate of working. The unit of power is called the 
Watt, and is equal to io 7 ergs per second. 

THE HORSE-POWER. 

1 foot = 30.479 centimetres. 

1 lb. =453 -59 grams. 

.. 1 ft. lb. =£-(30.479x453.59) = 13562600 ergs =1.35626 x 10 ergs 

One horse-power=33000 ft. pds. per minute= 55 o ft. pds. per second 
=550 x 1.35626 x io 7 ergs. 

But 1 Watt=io 7 ergs. 

1 H.P. = 550 x 1.35626=745.941 Watts 
=746 Watts very nearly. 

E 2 

H.P. x 746 =E C= C 2 R=-p 


h.p. = ec - 

746 


C*R 

746 


E 2 


746 R 


The unit of Quantity is called the Coulomb=lo 1 C GS units. It 
is the quantity given by an ampere in a second. 

1 volt Coulomb or 1 Watt during every second) =10,000,000 ergs. 

1 volt ampere during every second, or Joule j = -7373 2 4 f° ot P^ s * 
The Joule (Joule’s mechanical equivalent) is therefore equal to 
the work done or heat generated by a Watt in a second and is 
= .737324 ft. pds. 

.*. E.C.t=C 2 Rt=?^=EQ =Joules 

K 

when £)=quantity in Coulomb. 

Work in foot pds. = .737324 E Q 

Example—How much electricity will 330,000 foot pds. send 
through a circuit with an E.M.F. of 60 volts ? 

Work=. 737324 E Q 


Q=- 


330000 


-=7460 Coulombs. 


.737324 x 60 
Now Coulombs per second= Amperes 

7460 x 6o=Watts per second 
• 7460 x60 =HP ;^ j3o ! o® = feo horse-power. 

746 550 

Summary of Formulas— 

h.p.=£ 2 - 

746 
_ EX 
746 

_E*_ 

74 6 







engineers’ manual. 



I E. L. McALLASTER 


MECHANICAL ENGINEER 

and 

NAVAL ARCHITECT 


JL 503-504 PIONEER BUILDING 

~T T plmhnne Main 445 SEATTLE. WASH. 




136 


engineers’ manual. 


Work in foot pounds= .737324 EQ 


^ E V746 H.P. 
~~R~ R 
746 H.P. 


E 


Work in ft. pds. 


746 H.P. _- 

£=C.;?=^_- = y 74 6 H.P.R.- - r „ 7324 Q 


E 74 6 H.P. E*_ 
~CT~ C* “746 H P. 


THE HEATING EFFECT OF THE CURRENT. 

The unit of heat called the therm or French caloric is that quan¬ 
tity of heat necessary to raise 1 gram of water i° centigrade. 

The British unit = 1 lb. deg. F. = 772 ft. pds. 

= 1.403 H.P. 

But 1 H.P. = 746 Watts. 

746 x 1.403 = 1047.03 Watts for 1 B.H.U. 
but 1 Watt = io 7 ergs. 

1 lb. deg. F. = 1047.3 xio 7 = 1.0473X10 10 ergs. ' 

From this follows that 1 lb. deg. cent. = 1884.66 x io 7 ergs, 
and as there are 453 -59 grams per lb. 

1884.66 x io 7 ^ Tx io 7 er g- s . 

453-59 

or 1 gram. deg. centigrade = 4. 15495 x io 7 ergs. 

Let J = Joules mechanical equivalent. 

= amount of mechanical work 1 caloric is capable of 
doing. 

H — number of heat units. 

JH = work done = C 2 R.t. 
when t — time in seconds. 

If Q = the quantity of electricity passed 

E — E. M.F. or difference of electrical level, then, as in lifting a 
weight, the work done against gravity is mass x height through 
which it has been raised, so 

Q E = total work = W 
J H — Q E = W. 

But since C = the quantity that passes each second and t the 
number of seconds, then 

Ct = total quantity passed =Q 
JH= QE=CtE. 

By Ohm’s law C — 

and substituting CP for E we get 

JH = C* Rt 











137 


engineers’ manual. 

A. Lundberg 

ARTIFICIAL LIMB CO. 

ROOMS 13 and 14, SULLIVAN BUILDING. 

AUTHORIZED MANUFACTURER FOR U. S. 
GOVERNMENT. 


Deformity Apparatus 

Of Every Description 


TRUSSES 

CRUTCHES 

ELASTIC STOCKINGS 
SUSPENSORIES 

STOCKINGS FOR 
ARTIFICIAL LIMBS 

SHOULDER BRACES 
ROLLING CHAIRS, ETC, 


710-713 FIRST AVENUE 
Phone White 478 SEATTLE, WASH. 






engineers’ manual. 


138 

The value of J is given as follows : 

J - 4.1549S x r ° 7 er & s ‘ for 1 gram deg ' Cent ‘ 

= 1884.66 x io 7 ergs, for 1 lb. deg. cent. 

— 1047.03 x io 7 ergs, for 1 lb. deg. F. 

Example—A current of 20 amperes flowing through 10 Ohms, R , 
heats 20 lbs. water from 60 to 65° F. Find length of time, C, was 
flowing. 

C?Rt 

h= ~t 

H=20 lbs. (65-60) =100 units (lbs. deg. F.) 

J= 1047.03 x io 7 

The formula so far is in absolute units, and to reduce same to prac¬ 
tical units we have 


II 


(Cx io- 1 ) 2 x R x. io 9 x t C 2 Rt 


1047.03 X IO 7 

20 2 X IO Xt 

1047 

^=26.175 seconds. 


1047.03 


In this example no allowance has been made for radiation. 

Example—A current was sent through a wire of 12 Ohms resist¬ 
ance, wholly immersed in 25.5 grams of water, contained in a glass 
vessel. At the end of 4 mins, the rise in temperature was observed to 
be 30°C. Calculate the strength of the current. Ans.—1.05 amperes. 


COMPARISON OF HEAT. 

Suppose it is required to compare the amount of heat produced 
in wires of different resistances by currents of different strength for 
different times. 

Let the heat in one wir e=H 1 = C 2 Rtx .24 
“ “ 2nd “ =H 2 = C 2 2 ^2^2 x - 2 4 

H C^Rtx .24 C' 2 Rt 
C^R^t 2 x .24 C 2 t? 2 ^ 2 

That is to say, the heat produced in one wire, multiplied by product of 
current squared, resistance and time in seconds of second wire, is 
equal to the heat produced in the second wire multiplied by the 
current squared multiplied by the product of the current squared, 
resistance and time in seconds of the first wire. 

Example—The resistance of one wire is 5 Ohms, and that of 
another is 4 Ohms. Find the ratio of the heat produced in the one 
wire to that produced in the other wire—(1) when joined in series ; (2) 
when joined in multiple when a current is sent through them. Ans.— 
fi ) H x : H 2 :: 5 : 4; (2 ) H, : :: 4:5. 











engineers’ manual. 


1 39 


HEATING BY ELECTRICITY. 


It is found that the heat produced in a conductor is directly 
proportional—(i) to the square of the current; (2) to the resistance 
of the conductor; and (3) to the time the current is flowing or 
expressed by the equation. 

JH=C>Rt, 

when y=Joule’s mechanical equivalent. 

H= Number of heat units. 


As 1 H,P. =550 ft. lbs. =746 Watts and one British heat unit is equal 

* 7*72 

to 772 ft. lbs., therefore 1 lb. deg. F. is equal to ^^=1.403 H.P. or 

^47.3 Watts. This will represent the value of ywhen dealing with 
British heat units : 


C 2 Rt C.Et EH 
~ 1047 ~ 1047 _ 1047^? 

In the best of lighting and power plants it takes a coal con¬ 
sumption of 2\ lbs. per indicated horse-power per hour; and allowing 
90% efficiency in the engine, 93% in generator and 90% in the circuits, 
we get, say, 75% combined efficiency, or for every horse-power 
generated at the engine we get § H.P. at the heater on the con¬ 
sumer’s premises, which is equivalent to 3.3 lbs. coal per E.H.P. 


For a coal consumption of 2\ lbs. we 

C 2 Rt 

H = — t~ —1926 


or 


n — j — 1 yzu 

770 heat units per lb. coal. 


In good hot water or steam heating systems an average of 9500 
heat units are utilized per lb. coal. Therefore, the relative efficiencies 
are as 770 : 9500 or 1 : 12.5 ; that is to say, to heat by electricity 
would cost 12^ times more than by steam. As there are very few 
plants, generating one 1 H. P. for 2\ lbs. coal, this ratio is much higher, 
the majority of plants having a coal consumption of 6 lbs.; therefore, 
assuming 4 lbs. as the average, we get the relative efficiency as 
being 1 : 20. When very small quantities of heat are required and 
one momentarily, the electric heater is preferable and more economi¬ 
cal than anything else. 


Size of Wire Necessary to Carry a Given Current. 

Required the size of wire necessary to carry 30 amperes at dis¬ 
tance of 1300', allowing a loss of 5% at 100 volts. 

E E 

By Ohm’s law we have C = — or R = — 

J R C 

Applying this to finding the resistance of the wire we get R — 

— } Ohm the total resistance of the whole circuit or 2600'. \ Ohm for 
2600' = { x j| = .064 Ohms per 1000'. 






engineers’ manual. 

By referring- to the table given below we find that to be ooo wire. 
From the above we can deduce a formula, as Follows; 

L x 1000 
R = ,Vx 2 D 

where JR = resistance per iooo' 

L = loss in volts 
C — total current 
D = single distance. 

Another Example— Find size of wire necessary to carry 20 am¬ 
peres a distance of 5000', allowing a loss of 8%. Voltage 2000. 

^•-rhofMOO x ;°o° = 8Qhm 

20 X 1OOOO 

No. 9 wire according to table has .811 Ohms, therefore No. 8 
would be used. 

The above rule is good for any system and any voltage. 

TABLE OF RESISTANCES 


SIZES, WEIGHTS AND LENGTHS OF COPPER WIRE. 


Gauge 
No . 

Size. 

Weight and Length 

Resistance. * 

Carrying 
Capa¬ 
city, 2000 
Amperes 
per sq.in 

Diam. 
in Mils. 

3ia. 2 or 

Circular 

Mils. 

Pounds 

per 

1000 feet. 

Feet per 
Pound 

Ohms 

per 

1000 feet 

Feet per 
Ohm. . 

OOOO 

460.ooo 

211600.0 

639.60 

1.564 

.051 

19929 7 

43° 

OOO 

OO 

409.640 
364.800 

167804.9 

i33°79-° 

507 22 
402.25 

1.97 1 

2.486 

.063 

.080 

15804 9 

12534 2 

262 

208 

O 

324.950 

105592.5 

3x9.67 

3 *33 

.101 

994i•3 

165 

I 

289 300 

83694.5 

252.98 

3-952 

.127 

7882.8 

130 

2 

257.630 

66373.22 

200 63 

4-994 

.160 

6251 4 

103 

3 

229.420 

52633.53 

i59 °9 

6.285 

.202 

4957-3 

81 

r. 

4 

204.310 

41742 57 

126 17 

7-9 2 5 

.254 

393 £ -6 

65 

5 

181.940 

33102.16 

xoo 05 

9 995 

.321 

3 XI 7 8 

52 

6 

162.020 

26250.48 

79-34 

12 604 

• 404 

2472.4 

41 

7 

144.280 

20816.72 

^62.92 

15-893 

• 5°9 

i960 6 

32 

8 

128.490 

16509.68 

49 9° 

20 040 

• 643 

1555 0 

26 

9- 

114.430 

13094.22 

39 58 

25.265 

.811 

1233-3 

20 

10 

101.390 

10381.57 

3!-38 

31.867 

1.023 

977 8 

16 

ii 

90.742 

8234.11 

24 89 

40 176 

1.289 

775-5 

" *3 

12 

80 808 

6529.93 

19 74 

5° 659 

1.626 

615 02 

10 2 

13 

71.961 

517 s -39 

15-65 

63 898 

2.048 

488 25 

8 1 

0 

14 

64 084 

4106.75 

12.41 

80.580 

2.585 

386 80 

6 4 

15 

57.068 

3256.76 

9.84 

101 626 

3-177 

306 74 

5 1 

16 

50 820 

2582.67 

7.81 

128 041 

4.582 

243 25 

4 0 

17 

45-257 

2048.19 

6.19 

161 551 

5 vi 83 

192 91 

3 2 

18 

40.303 

1624.33 

3.786 

203 666 

6.536 

152.99 

2 5 

19 

35-39° 

1252.45 

3.086 

264.136 

8.477 

117 96 

1.96 

20 

31 961 

X021.51 

2.448 

324.045 

io-394 

96 21 

1 60 

21 

28 462 

810.09 

x-94 2 

408.497 

13.106 

76 30 

1 28 

22 

25-347 

642.47 

1-539 

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16.*25 

■ 60 51 

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1.221 

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20.842 

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41.789 

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52.687 

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engineers’ manual. 


141 

THE CHEniCAL EFFECT. 

If we were told that a certain quantity of water—say, 100 
gallons—had passed through a pipe, this by itself does not give us 
any idea of the force of the flow, or in an electrical sense the strength 
of the current. It might have taken a week to trickle through, or it 
might have passed in one minute ; and according as the time is short 
or long, so is the force of the flow greater or less. 

We must not only know the total quantity that has passed, but 
the time taken in its passage must also be known, to get a definite 
notion of the strength of the current. The current is the quantity of 
electricity that passes any part of the circuit in unit time, i.e. , one 
second, and the unit of quantity is called the Coulomb ; the practical 
unit of current is called the Ampere, and coulomb per second=amperes. 

The amount of chemical action at all points of the circuit are equal 
to one another. This does not mean that the same current passing for 
the same length of time through different solutions will decompose 
equal weights of the metals contained in these solutions, but that the 
weights of the metals so decomposed will be chemically equal, i.e., 
the weight will be in direct proportion to the chemical equivalent. 

The electro-chemical equivalent is the weight of a substance 
decomposed by the passage of one coulomb. 

Let M= total mass in grams decomposed. 

_y=mass decomposed by 1 coulomb in grams=electro- 
chemical equivalent. 

^=time in seconds. 

£ 7 =current in amperes. 

. M 

Formula, C— —r 

yt 

.'. M—C.y.t. 

From this, we can calculate the consumption of zinc in a battery 
where the value of y—. 000337. The weight consumed per cell = 
Cty=Ct. 000337 grams per second, which is=Cxi.2i3 grams per 
hour. If there are n cells the total weight of zinc consumed is= 

nCty nCt. 000337 ... . 

Hi— -—jr-r = i r=— ; -77-7=weight in grams per second. 

No. in parallel No. in parallel b 

The Chemical Effect 

N. Cxi .213 

or Total weight consumed = No. in parallel. = & rams P er hour - 

Total weight consumed __ N.Cx 1.213 __ N. C x .00267 4 

in lbs. per hour. ~ No. in parallel x 453.6 No. in parallel. 

If all cells were in series, then the total weight consumed in lbs. 
per hour = 

JV=n.Cx .00674, and suppose we have a current of 746 am¬ 
peres at 1 volt which is = 1 horse-power, and substitute this value for 
n C we get 

W = 746 x . 002674 = 2 lbs. zinc at 1 volt. 

Therefore for a higher E. M.F. the consumption of zinc would be 
inversely proportional to the E. M.F. or 

H.P. per hour x 2 

Weight of zinc in lbs. = eTmTL 









142 


engineers’ manual. 


By means of the following- table the amounts deposited can be 
calculated when the current strength together with the tune are known. 


TABLE OF ELECTRO-CHEMICAL EQUIVALENTS. 


Elements. 

Valencies. 

A tomic 
Weight. 

Chemical 

Equivalent 

Electro-Chem. 
Equiv in Grams 
per Coulomb. 

Aluminium. 

3 

27-3 

9.1 

.00009449 

Gold. 

3 

196.6 

65-5 

.0006780 

Silver. 

1 

108. 

108. 

.0011180 

Copper . 

2 

63 - 

31 -5 

.0003261 

Tin.. , . 

4 

118. 

29.5 

.0003054 

Nickel. 

T 

2 

58.6 

29-3 

.0003054 

Zinc. 

2 

65 - 

32-5 

.0003364 

Hydrogen. 

1 

1 . 

1 . 

.000010352 


The atomic weight is the weight of the atom, the weight of an 
atom of hydrogen being taken as 1. The atomic weight of copper is 
63, i.e. 63 times heavier than hydrogen ; but in chemical combination 
one atom of copper replaces 2 of hydrogen, hence the weight equiva¬ 
lent to 1 of hydrogen is 63 4 - 2 = 31.5. Therefore the atomic weight 
-r- valency is = the chemical equivalent. 

Example—A current of 2.5 amperes passes through a solution of 
gold for 20 minutes. What will be the total deposit? 

According to the table the electro-chemical equivalent is = 

.0006780. 

/. M — Cyt = 2.5 x .0006780 x 20 x 60 
= 2.034 grams. 

Example—How long would it take to silverplate 6 spoons sup¬ 
posing the current was 1.5 amperes and that each spoon would take 
. 125 grams of silver to cover it. Ans.— 7 minutes nearly. 

DYNAMO=ELECTRIC MACHINERY. 

A dynamo-electric machine is a machine for converting energy 
in the form of mechanical power into energy in the form of electric 
currents, or vice versa , by the operation of setting conductors, 
usually in the form of coils of copper wire, to rotate in a magnetic 
field. (See Sylvanus P. Thomson’s “ Dynamo-Electric Machinery.”) 

Faraday in 1831 made the discovery that, by moving conductors 
in a magnetic field, electric currents are generated in them, and 
the principle of magneto-electric induction is as follows : 

When a conductor is moved in a field of magnetic force so as to 
cut the lines of force, there is an E.M F. produced in the conductor 
in a direction at right angles to the direction of motion, and at right 
angles also to the direction of the lines of force, and to the right of 
the lines of force, as viewed from the point from which the motion 
originates. 

The induced E.M.F. is proportional to the number of lines of 
force cut per second, ^.nd is therefore proportional to the intensity of 
the “ field” and to the length and velocity of the moving conductors. 





























engineers’ manual. 


143 

As the volt is equal to io 8 C.G.S. units, then the number of volts 
generated by a rotating- armature is 

E—Revs. per second x No. of conductors in series around arm¬ 
ature = Total lines of force which pass through the armature 
core, divided by io 8 . 
or R.p.s x No. of cond. x Flux 
100,000,000 

By this we see that, to increase the E.M.F., we can do so by 
increasing the speed, or increasing the number of conductors, or 
increasing the lines of force, or all three of them could be increased. 

The dynamo consists of two essential parts, viz., the field magnet 
and the armature. In the majority of continuous current machines 
the revolving part is the armature, and the field magnets are 
stationary. 

There are several methods of exciting the fields, viz., by per¬ 
manent magnets or electro magnets, self-excited or otherwise. Hence, 
the current of the generator may be itself utilized to excite the mag¬ 
netism of the fields by being caused, wholly or partially, to flow 
round the field windings. 

In the shunt wound dynamo, the field magnet is wound with a 
large number of turns of fine wire, and, being in shunt with the main 
current, only part of the whole current generated in the armature. 
The shunt machine is less liable to reverse its polarity than the series 
dynamo, and may be controlled so as to give a uniform E.M.F. by 
introducing a variable resistance into the shunt or field circuit. 

When a shunt machine is supplying lamps in parallel, the turning 
on of additional lights reduces the resistance of the circuit and in¬ 
creases the current, but not in proportion, for when the resistance of 
the main circuit is lowered a little less current flows around the field 
windings and lessens the magnetism. 

The series-wound dynamo consists of but one circuit. The 
majority of arc machines are series wound. The whole current from 
the armature is carried through the field windings, which are. in 
series with the main circuit. 

Any increase in the resistance of the series-wound dynamo less¬ 
ens its power to supply current, because it diminishes the flux. When 
lamps are in series (as in the ordinary arc lighting), the switching on 
of an additional lamp both adds to the resistance of the circuit and 
diminishes the power to supply current. It requires the same expen¬ 
diture of energy to magnetize an electro-magnet to the same degree 
whether shunt or series wound. 

We see from the above that in the shunt-wound dynamo by turn¬ 
ing on more lamps the E.M.F. is reduced, and in a series machine the 
switching on of additional lights, if in multiple, will increase the 
E.M.F., and by properly proportioning the series winding and com¬ 
bining the two windings we could get a steady E.M.F. This is 
exactly what is done in the compound-wound machines. 

ARflATURES. 

If ii on is employed in armatures it must be laminated so as to 
prevent Foucault current. Cores built up of varnished iron wire or 
of thin discs of sheet iron separated by varnish or paper realize this 
condition. 



i 4 4 


engineers’ manual. 


All needless resistance should be avoided in the armature coils, 
as hurtful to the efficiency of the machine. The wire therefore should 
be as short and as thick as is consistent with obtaining the requisite 
E.M.F. without requiring an undue speed of driving. 

Since it is impossible to reduce the resistance of the armature 
coils to zero, it is impossible to prevent heat being developed in those 
coils while the machine is generating currents. 

The insulation of the armature should be insured with particular 
care, and especially at the ends in drum wound armatures, where 
there are numerous crosses. 

COMMUTATORS. 

Approach being a finite process, the method of a coil approach¬ 
ing and receding from a magnet pole must necessarily yield currents 
alternating in direction. By using a suitable commutator, all the 
currents, direct or inverse, produced during recession or approach, 
can be turned into the same direction in the wire that goes to supply 
currents to the external circuits; and if the rotating coils are properly 
grouped so that before the E.M.F. in one set has died down, 
another set is coming into action, then it will be possible, by using an 
appropriate commutator, to combine their separate currents into one 
practically uniform current .—Sylvanus P. Thomson. 

The commutator in direct current machines is the most trouble¬ 
some part of the whole machine, and great attention should be paid 
to it, to have the brushes bearing at the proper angle, and in a bi- 
poler machine set diametrically opposite. 

THE NEUTRAL POINTS. 

In consequence of the armature itself when traversed by the cur¬ 
rents, acting as a magnet, the lines of force will not run straight 
across from pole to pole, but will on the whole assume an angular 
position, being twisted in the direction of rotation a considerable 
number of degrees. Hence the diameter of commutation, which is at 
right angles to the resultant lines of force, will be moved forward. In 
other words, the brushes will have a certain angular lead ; this lead 
depending upon the relation between the intensity of the field and the 
current in the armature. 

Hence, in all dynamos, it is advisable to have an adjustment, 
enabling the brushes to be rotated round the commutator or collector 
to the position of the diameter of commutation for the time being. 
If this is not done, there will be sparking at the brushes, and in part 
of the coils at least the current will be wasting itself by running 
against an opposing E.M.F. 

EFFICIENCY. 

The efficiency of a dynamo-electric machine is the ratio of the 
useful electrical work done by the machine to the total mechanical 
work applied in driving it, Every circumstance which contributes to 
wasting the energy of the current reduces the efficiency of the 
machine. 

^Electrical loss cannot be obviated entirely, because the very 


engineers’ manual. 


'45 

oest of conductors have some resistance. Mechanical friction of the 
moving- parts should be brought to a minimum. 

~ f j Internal Elect. H.P. 

I he mechanical efficiency of a dynamo —- t v v ; . —:—:- 

H.P. in belt. 

, ,, t 4. j „ n E. M.F. in armature x arm. current 

when the Internal Electrical H.P.= ---- 

746 

t ™ r , External Electrical H.P. 

Electrical efficiency of a dynamo = - Total Electrical H P whenthe 

^ , T , t . , TT „ E. M. F. atterminals x External current 

External Electrical H.P.=-7—- 

746 

and the total Electrical H.P.=poweractually converted inthearmature. 

Thus, if it took, say, 3% of the total E.H.F. for the field winding 
and other 3% was wasted in heating the conductors of the armature, 
then the electrical efficiency will be 94% of the gross electric power. 

_ . , „ . „ External Electrical H.P. 

Commercial efficiency ot a dynamo =- H P in belt - 

In good machines this reaches higher than 90%, while the electrical 
efficiency is as high as 97 %. Horse-power in belt=gross indicated 
horse-power of engine - engine friction. Deduct, roughly, 15% of 
mean pressure for friction. 

The following has been selected from the instructions issued some 
years ago by the Edison Electric Co., and may be taken as fully 
covering the ground : 

Location, Setting and Starting of Dynamos. 

The dynamo should be located in a clean, dry place, and prefer¬ 
ably in a room of low temperature. 

The foundations should be of a substantial character, solid mason 
work or stout framing, sufficient to obviate all vibration while the 
machine is in operation. 

The proper insulation of the dynamo from “ earth ” is vital. To 
secure this a stout frame of heavy timber is provided ; this is secured 
to the foundation. The frame should be thoroughly treated with 
some moisture repellant such as asphalt varnish. 

Exercise great care in handling the armature. Use only rope 
slings and wooden bars. 

Handle as much as possible by the shaft. 

Never, under any circumstances or in any manner, make use of 
the commutator in handling the armature. 

Do not allow the weight of the armature to rest on it for a moment. 
Never lay an armature down unless you have a thick, soft pad 
between it and the floor. 

It is quite important that before a new dynamo is put at steady 
work it should be run for a few hours first at slow speed, which may 
be gradually increased to the maximum. During this trial run, care¬ 
fully attend to the bearings. Make sure that everything is in perfect 
condition previous to putting the dynamo at work on the circuit. 









engineers’ manual. 


146 

Cleanliness about the Dynamo. 

All parts of the dynamo should be kept neat and clean. Dirt, 
copper dust and oil should not under any circumstance be allowed to 
gather on any part. 

Never allow loose articles of any kind to be placed upon any por¬ 
tion of the dynamo. 

Adjustment of Brushes. 

In order to maintain the commutator in proper condition and 
reduce the wear to a minimum, it is vitally necessary that a proper 
adjustment of the brushes be secured. They should work absolutely 
free from sparks. Any sparking whatever indicates a bad condition 
of the commutators or defective adjustment of the brushes. 

The end of the brush should be carefully bevelled so as to con¬ 
form accurately with the surface of the commutator. The brush 
should bear lightly upon the commutator, and every part of the 
bevelled end should rest upon it. The pressure should be just suffi¬ 
cient to ensure good contact, and avoid all cutting and scratching. 

One of the worst causes of sparking is lack of pressure of the 
brush, caused by improper setting of the brush-holder stud or by 
allowing a brush to wear too short. To maintain the proper angle, 
the brush as it wears must be pushed forward in the holder from 
time to time. 

A dynamo in operation with sparking brushes is prima facie 
evidence of carelessness or ignorance on the part of the attendant, 
and such a condition of affairs should not be tolerated under any 
circumstances. 

Causes of Sparking. 

Brushes not set at neutral point. 

Brushes not set at diametrically opposite points. 

Brushes set so as not to get full bevel to the circumference of 
commutator. 

Brushes set with insufficient pressure. 

Brushes spread apart and filled with oil and dirt. 

Commutator bars loose, high or low. 

Loose connection between armature coil and commutator bar. 

Section short circuited either in commutator or armature coils. 

Armature damp, with consequent short circuiting of coils. 

Short circuit or cross on outside system. 

Commutator dirty, oily, rough worn in ridges, or out of truth. 

Dynamo overloaded. 

Armature coils or commutator sections short circuited by accu¬ 
mulation of copper dust. 

N.B .—An examination of some dynamos would lead one to believe 
the machine was constructed for the purpose of producing copper 
dust. The accumulation of copper dust on a dynamo, and its gradual 
penetration into the armature and field coils, is often the real cause 
of serious accident and expensive repairs. This is one of the prin¬ 
cipal features which denotes carelessness and inefficient manage¬ 
ment, and an utter lack of appreciation of the importance of 
cleanliness about dynamos and electrical apparatus. The remedy is 
easy to apply ; the dynamos must be kept clean of oil and copper dust. 


ENGINEERS’ manual. 


-+7 

The following are some of the disorders which dynamos are 
subject to :— 

Burning Out an Armature Coil. —This may be occasioned by 
overloading the armature, causing the insulation of the coils to give 
way, and is indicated by the armature suddenly beginning to smoke. 
The coil is thus rendered useless. As a temporary make-shift, the 
injured coil may be disconnected from the commutator, the ends insu¬ 
lated with tape and the two adjacent bars to which the coil was con¬ 
nected joined to each other by a wire not less than the armature wire. 
The machine can be operated for a time in this way, but it should be 
repaired at the first opportunity. 

Ring of Fire Around the Commutator. —This is caused by small 
particles of copper between the bars of the commutator, making a 
local short circuit from bar to bar across the mica insulation. Clean 
the commutator carefully and do not allow the brushes to cut and 
scratch it. 

Breaking Down of one Dynamo. —If one dynamo of a single pair con¬ 
nected in series on a 3-wire system breaks down, the result will be 
merely to put out the lights on that side of the system. If, however, 
otl er machines are in multiple with the disabled one, the current 
through the armature will be reversed, and if not disabled electrically 
will run as a motor. Cut the machine out at once. 

Reversal of Polarity of Magnets. —Reversal of polarity of a dynamo 
which is one of tw r o or more connected in multiple is equivalent to a 
dead short circuit and if it does not blow the fuses or circuit breaker, 
or throw off the belt, will probably burn the armature out. 

Reversal of polarity of one of a single pair of dynamos working 
in series on a 3-wire system, will tend to send all the current 
through the central wire, which will cause the lights to burn dim. 

More trouble will be caused by switching in a reversed machine 
with another that is not reversed. 

Dynamos on the 3-wire system may be reversed under the follow¬ 
ing conditions : 

A reversal sometimes occurs when starting up, caused by the 
influence of another dynamo in close proximity to it. 

Reversal may be caused by the current of the second dynamo in 
series while in operation, if the brushes of the first dynamo are raised 
or its current broken in any way between the points to which the 
field circuit is connected. 

By lifting the brushes before throwing out the switch. 

By burning out the safety catches on some other dynamos. 

By crosses on the line. 

By 200-volt motors. This is more apt to occur during a light load 
when the motor is thrown on with a heavy load. 

To correct the polarity, open the circuit switch, raise the brushes, 
throw in dynamo switch on the side not reversed and leave it about a 
minute. 

Effects of Lightning. 

One of the safest places to be in during a thunder-storm is in an 
electric light station. 


148 


ENGINEERS MANUAL. 


In underground systems no effects of lightning are felt, but 
where there are long out door pole lines the same effects occur as on 
telegraph and telephone lines, and certain precautions must be taken 
to prevent injury to apparatus. 

Lightning arrests should be in plain sight. Fuses on such 
arresters must be promptly replaced, and ground wires and connec¬ 
tions must be kept intact and in good condition. 

Facts to be Remembered. 

Be sure that the speed of the dynamo is right. 

Be sure that all belts are sufficiently tight. 

Be sure that all connections are firm and make good contact. 

Keep every part of the machine and dynamo room scrupulously 
clean. 

Keep all the insulations free from metal, dust and gritty sub¬ 
stances. 

Don’t allow the circuit to become uninsulated in any way. 

Keep all bearings of the machine well oiled. 

Keep the brushes properly set, and see that they do not cut or 
scratch the Commutator. 

If brushes spark, locate the trouble and rectify it at once. 

Before throwing dynamos in circuit with others running in 
parallel, be sure the pressure is the same as that of the circuit, then 
close the switch. 

Be sure each dynamo in circuit is so regulated as to have its 
full share of the load, and keep it so. 

ELECTRIC MOTORS. 

Any dynamo can be run as a motor, and the instructions given 
above regarding dynamos are applicable to motors. 

In the dynamo only one E M.F. exists, whereas in the motor 
there must be two, viz., the E.M.F. of supply and the counter E.M.F. 

In the dynamo the current flowing through the armature is= ^ 

where R is the total resistance of the circuit. With the motor the 
E.M.F. of supply - counter E.M.F. 
current flowing is= - ^ --—- where R y is 

the resistance of the armature. 

E 200 volts 

5 kilowatt dynamo C—^ 25 amperes =; ~g ' ms 

.. _ E.M.F. - coun. E.M.F. 200-180 

5 motor C= - ^ - =25 amperes = - g— 

From this, we see that the current and the E.M.F. is the same in 
both cases, but the resistance of the motor circuit is one-tenth that 
of the dynamo, the difference being made up by the counter E.M.F., 
which has the same effect as resistance. The ratio of the E.M.F. of 
supply to the counter E.M.F. is the electrical efficiency of the motor. 








ENGINEERS’ manual. 


149 


]ourne & Clark Co, 


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815-817 Second Ave. 





150 


ENGINEERS’ MANUAL. 


Pacific Metal Works 

IMPORTERS OF 

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AND MANUFACTURERS OF 

Babbits Solder and Metal 
« « Alloy « « 

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INDEX 


Page 

Acceleration Due to Gravity.._....ioo 

Algebra.. 15 

Ammonia, Table of Properties. 113 

Areas of Circles. 51 

Areas of Segments of a Circle .... 41 

Arithmetic. 11 

Arithmetical Progression. 23 

Armatures........ 143 

Belting..„...... 67 

Belts, Horse-power Transmitted by. 116 

Boilers, Care of. 81 

Boilers, Foaming in. 59 

Boiler Settings. 85 

Boiler Shells, Strength of..... 107 

Brake, Horse-power....... 115 

Chemical Effect, The.... 141 

Circumference of Circles. . 55 

Commutators... 144 

Current, Division of. 132 

Current, Heating of..... . 136 

Decimals.- . 13 

Dynamo—Electric Machinery... 142 

Dynamos, Causes of Sparking......... 146 

Dynamos, Location, Setting and Starting- 145 

Effects of Lightning and Facts to bo Remembered- 147 

Electric Motors.. 148 

Electrical Units of Measurement. 132 

Electricity.-. 128 

Electricity, Heating by. 139 

Electro-Chemical Equivalents, Table of.. 142 

Engine, Duty of an. ug 


































152 


ENGINEERS 1 MANUAL. 


(•XjX^XsXjXiXsXsXsXjXsXjXsXSi®® ®®®®(*)(?X*X*X*X5X5(5X£X»X*X»)(5X5X£)®@®®®®0 




(!) 


BELT CEMENT, 

BELT DRESSING, 

BELTING CLEANED, 
BELTING REPAIRED. 

rnTTTTT 

R.B.TOLSMASCO. 

BELT MAKERS 

DEALERS 

Leather Belting, Rubber Belting 
and Lace Leather 

TELEPHONE MAIN 78 


309 OCCIDENTAL AVENUE, 

SEATTLE, WASH. 






ENGINEERS* MANUAL. 


Engine, How to Set Up a Stationary.... 

Engine, To Place on.the Dead .Center.*. j _. .. 

Evaporative Tests _ .. . . .. .. 

Evolution _ ___ 

Feedwater, Heating of.. .. . . . 

Flywheel, Energy and Bursting Stress . . 

Fractions. . . 

Geometrical Progression ... j . _~j 

Geometry, Practical...._I__ j.... ___ i, .1 _ 

Horse-power, Rules fpr Determining_ . . . . 114- 

Injeetbr, The . ...i _ 

Mean Pressure, Graphic Method cf Finding _ 

Mechanical Refrigeration and Ice Makings _ 

Mechanical Stoker J___ 

Mensuration of Solids .... 

Mensuration of Surfaces. __...___ 

“Of Interest to Engineers” ... 

Pendulum, The_ __ 

Polygons, Table of Regular ___ 

Pulleys, Rules for Calculating Size and Speed . 

Pumps. ..... 

Quadratic Equations__ _ 

Resistance, Derived Circuits of _ _ 

Resistances, Table of _ r _ __ _ 

Rivetted Joints.... . 

Safety Valve, The . ...1 __ _ 

Screw Cutting... . .. 

Shafting, Strength of Solid Round. _ _ 

Shafting, Power Transmitted by ___ 

Stays .... . ... 

Steam. .. . .... 

Steam, Expansion of.. . 

Steam, Properties of Saturated _ 

Useful Information_ _ _ 

Valve Setting____ 

Valve Setting Duplex Pump--- 

Water at Different Temperatures_ 

Wire, Size Necessary to Carry a Given Current 
Zeurner’s Diagram. . 


101 

121 

85 

2b 

90 

117 

II 

25 

29 

■134 

89 

80 

III 

38 

49 

29 

124 

97 

35 

95 

9 1 

23 

130 

140 

103 

122 

96 

93 

94 

108 

7i 

75 

73 

65 

120 

93 

109 

139 

119 









































154 


engineers’ manual. 


* * Patents Guaranteed * * 


Nason, Fenwick 8 lawrence 

PATENT LAWYERS 
AND EXPERTS 


Oldest Patent Law Firm in Washington, D. C. 

40 Years’ experience in attending to and securing 

PATENTS 

Caveats, Copyrights and Trade Marks 


Infringements and appeals vigorously prosecuted. 
Practice in all courts. Branch office in charge of 
G. WARD KEMP, Attorney-at-Law. 

432 BURKE BLDG., SEATTLE, WASH. 


Same Service and Charges in Branch Offices. 
Free Guide Book. 






engineers’ manual. 


155 


INDEX TO ADVERTISEMENTS. 


Adair & Son, Geo. R. 

Adams, Frank E. 

Allmond, Chas. Ii .. 

American Steel & Wire Co.... 

Baker & Richards . 

Beverleigh Mfg Co. 

California Saw Works .... . 

Campbell & Sons, Wm.. 

Canadian Bank of Commerce, The 

Clark & Miller . VVVVV. VV. ' .‘ VVV. V. ' 

Commercial Street Boiler Works, H. W. Markey, Prop 

Denny Clay Co. 

De Solla, Deussing Co .s 

Donaldson Brothers . 

Engineers Supply Co.. 

Felitz Tent & Awning Co. 

Finn Metal Works, John . 

Graton & Knight Mfg. Co. 

Griffing Iron Co., A. A. 

Heffernan Engine Works. 

Henshaw, Bulkley & Co. 

Houghton, E. W. 

Issaquah Coal Co.• 

Ivilbourne & Clark Co. 

Lundberg, A.. 

McAllaster, E. L. 

Mason, Fenwick & Lawrence . 

Mitchell, Lewis & Staver Co. 

Moore & Co., C. C.. . . . 

Northwest Fixture Co. 

Oregon Boiler Works, The .. 

Pacific Coast Co.,-The .. 

Pacific Heat & Power Co.-. 

Pacific Metal Works . 

Parker, Frank L.. 

Pierre Barnes & Co.. 

Tuget Sound Mach. Depot. 

Puget Sound Metal Works . 

Schram Co., John, The .. 

Seattle Bridge Co. 

Seattle Electric Co., The . 

Simonds Mfg. Co.. 

Soderberg Pipe Co.. 

Stumer, H. E.:. 

Sullivan Brothers . 

Sweeney & Ryan . 

Tolsma & Co., R. B. 

Tousey & Co., A. C. 

Union Brass Foundry . 

Washington Electric Supply & Mantel Co. 

Washington Iron Works Co. 


60 

12 

30 

118A 

24 

162 

16 

40 

161 

22 

26 

28 

62 

32 

46 

98A 

52 

68 

125 
118B 

64 

133 

54 

149 
137 
135 
154 

II 

I 

129 

36 

48 

44 

150 
18 
76 

6 

34 

20 

58 

126 
10 

8 

14 

42 

50 

152 

66 

38 

131 

56 


N. B. Please remember the advertisers when purchasing 
machinery and supplies, as it is through them we get this man¬ 
ual free of charge, and they are deserving of our support and 
patronage. 


INTERNATIONAL UNION STEAM ENGINEERS. 
Seattle Local, No. 40. 





















































ENGINEERS’ DIRECTORY 


Airey, Robt, 316 Minor Ave N. 

Albrecht, Frederick E, Albrecht Bros. 

Allen, Louis A, 191814 Fifth ave. 

Anderson, Jno R, 916 Ninth ave. 

Aukland, Harry, 514 Eighth ave. 

Auckland, Jno C, S W & T Co. 

Ayer Frederick, 1220 Valley. 

Baird, Amsey J, 215 Thomas. 

Barnard, Barney, 60S Fourth ave. 

Barrett, Wm L, Occidental ave, n e cor Second. 
Bassett, Chas, Elmer and Horricon, Edge water. 

Baumgart, Henry J, P C Co coal bunkers. 

Beckman, Fred F, Colfax, nr Pacific. S. Seattle. 

Beinsen, Herman A, Squire, nr sloop, Ballard. 

Belding, Emerson C, 816 Twenty-third ave. 

Bell, Harry, SATE Co. 

Bellamy, Guy L, Ballard. 

Benecke, Wm, H B Brewing Co. 

Benjamin, Wm, 810 King. 

Berggren, Fred, Seattle B & M Co. 

Bernsen, Herman A, N A T & T Co. 

Bethal, Walter, S B & M Co. 

Beyer, Otto, Washington B & Y Co. 

Biegert, Joseph C, Power House S C Ry Co. 

Bigelow, Miles, 1234 Lake View ave. 

Billingham, Henry, S C L M Co. 

Billingham, Thomas, S E Co. 

Bingham, Florian A, Hotel Seattle. 

Bird, Frank W, 675 Dearborn. 

Rolstad Gunerius, L, S & B Mill Co. 

Bourne, Nelson L, 811 Bell. 

Bowen, Chas A. Superior C & C Co. 

Bronson, Carl W, 1701 Second ave N. 

Brown, Arthur E, 2227 Third ave. 

Brown, Harry, Palmer House. 

Brulette, Wilson O, 819 Washington. 

Brulette, WmO, PSB&D Co. 

Burke, Jas E, 117 Lenora. 

Bye, Jonas L, Seattle Ice Co. 

Campbell, Wm, 1403 Seventeenth ave. 

Card, Jno E, Bryant L & S M Co. 

Carney, Patrick A, Edgewater, Bowman and Kilbourne 
Carroll, Michael J, 107 Seventh ave N. 

Case, Wm S, 4035 Brooklyn ave. Brooklyn. 

Casebere, Warner A, 2743 E Madison. 

Cashman, B J, Latona, Wash. 

Cedar, Carl R, 1218 Howell. 

. Cheney, Arthur, Rainier-Grand Hotel. 

Christof, Nels A, City Water Wks. 

Churchill, Jas E, I6I6V2 Seventh ave. 

Clemmance Wm F, Post st sta, S E Co. 

Colbert, Thos F, Hotel Seattle. 

Cole, Archibald A, Seattle Athletic Club. 

Colon, Albert E, 2908 Western ave. 

Conway, Edward L, N Y Blk. 

Cooper, Frederick, 126 Nineteenth ave n. 

Cooper, Herbert L, 1419 Queen Anne. 

Cooper, Jno W, P S B & D Co 
Cox, Landon A, Seattle C L M Co. 

Cree, Duncan H, 1328% 3rd ave. 

Crocker, Ebben F, The Arlington. 

Crowe, Edward H, 330 Fourth ave N. 

Currie, Jno, Heffernan Engine Works. 

Cutler, Albert R S, Seattle Cereal Co. 

Dagneau, Philip E, Campbell Mill. 

Dalby, Martin O, 908 Fourth ave. 

Daamon, Jno L, 512 Baily Bldg. 


aves. 


engineers’ manual. 


157 


Darlington, Wm T, 907 Blewett, Fremont. 

Darville, Ralph M, Fifteenth ave W, cor Atlanta, Smith's Cove. 
Darville, Wm, S M Co, Ballard. 

Davis, Grant T, 518 Boren ave N. 

Davis, Jno, 1905 Twenty-fourth ave S. 

De Hasseth, G. Andries, S E So. 

Denny, Henry L, Denny ave, Green Lake. 

Dewey, Phinney F, 1905 1st ave. 

Dickinson, Walter A, Globe Laundry. 

Dilg, Julius, Chlopeck Fish Co. 

Dimock, Arthur II, City Hall. 

Dixon, Jas R, 111 Seventh ave S. 

Dodds, Ed D, 2409 Western ave. 

Doolittle, Cornelius J, 1818 E Fir. 

Drennan, C B, The Richelieu. 

Duttenhoefer, Jacob, University of Washington. 

Duncan, Rob S, 120 Fifth ave s. 

Eckloff, Jno, Hotel Butler. 

Egar, Wm, Denion ave, Ballard. 

Ester, Chas, Alaska Com Hotel. 

Ettinger, Jno, Occidental Hotel. 

Evans, John A. S., B & M Co. 

Faber, Peter E., Diamond Ice Co. 

Fagan, Jno, 2306 Second ave. 

Faust, Adolph, 1123 Fourth ave. 

Fendall, Frank S, Brice’s Sta, Green Lake. 

Fisher, Jno O, W C Hill Brick Co. 

Fowler, Walter P, 807 Columbia. 

Fraser, Wm, 568 John. 

Frenzel, Chas, Arlington Hotel. 

Friend, Arthur, Van Asselt. 

Frost, Jos H, 604 Seventh ave S. 

Gade, Jno C, 400 Eighth ave. 

Gado, Jno S, B & M Co. 

Gagne, Philip, 338 Wilbert, Ballard. 

Gamble, Stephen A, 523% Pine. 

Gardener, Harry, 606 Sixth ave So. 

Garing, Albert C, 1324 Eighth ave So. 

Gelinan, South Park. 

Gillenwater, Eugene C, 1525 Sixth ave. 

Gilluly, Jas H, 319 Seventh ave So. 

Goebles, Daniel, North and Forth ave, Ballard. 

Goodwin, Frank L, P C Co. 

Griggs, Jno W, Etruria and Victory, Ross. 

Grogan, Cornelius, A, P C Co. 

Hagler, Lee, Alcatraz Asphalt Pav Co, 3035 Western ave. 
Hagstrom, Frederick F, King Mill Co. 

Hamblet, Alonzo, Ballard. 

Hampson, Wm B. Hester ave, Latona. 

Haneberg, Louis, 2119% Western ave. 

Hanford, II Day, S E Co. 

Hannah, Robert J, 323 Washington. 

Hannah, Wm, 323 Washington. 

Farbaugh, Harvey L, Post St Sta, S E Co. 

Hartman, Michael, S B & M Co. 

Harvey, Jno Novelty Mill, W Seattle. 

Ilayfort, C E, S B & M Co. 

Hansel, Frank J, Salmon Bay Shingle Co. 

Henry, H A, S Seattle. 

Henskey, Herman, 20 First ave and Grand Boul, Interbay. 

Hiatt, J G, 2228 Seventh ave. 

Hicks, Albert, 310 Thomas. 

Hill, Wm, First ave So and Conn. 

Hinckley, Chas, King Co Hospital, Duwamish. 

Hinckley, Lyman, Hinckley Blk. 

Hine, Homer II, SATE Co. 

Hodgson, Thos P, Seventh ave and Highland Drive. 


158 


engineers’ manual. 


Ilodson, F R, S Seattle. 

Hopkins, Burows H, Haller Bldg. 

Houtaker, Joseph, Seattle Ice Co. 

Hudson, Frank S, Seattle. 

Hufraan, Frank M, S E Co. 

Hurley, Frank, Madison House. 

Jackling, ffm B, 420 Seneca. 

James, Jno L, Third ave, Ballard. 

Jensen, Ole C, 941 Twtnty-third ave. 

Jodrell, Frank W, Pacific Coast Biscuit Co. 

Johnson, A, S Seattle. 

Johns, Wm, Seattle B & M Co. 

Johnson, Jno R, Fremont Barrel Mnfg Co. 

Johnson, John C, Ballard Steam Laundry. 

Jones, Chas H, Palace Laundry. 

Jones, James W, Fourteenth ave W. 

Kirk, P, South Park. 

Krist, Louis H, Carstens Bros. 

Kerry, Arthur E, Kerry Mill Co. 

Ivenzovich, Marco, Rapid Transit Co. 

Kilbourne, Edgar C, Chester, Brooklyn. 

Klaunig, Wm, 2207 E Cherry. 

Klinefelter, Geo W, 601 University. 

Knox, Harvey E, 1632 Seventh ave. 

Kroli, Albert R, Abrahamson & Co. 

Lame, Loemis C, 1918 First ave. 

Lancaster, Harry J, 75 Stewart. 

Landerkin, Geo M, Byron Sta. 

Larson, Andrew W, C M & I Co. 

Latham, Albert D, Roy & Roy Mill Co. 

Leigh, B H, Diamond Ice Co. 

Leigh, H Ii, Diamond Ice Co. 

Lindsay, Casper W, 1811% Seventh ave. 

Lindsay, J C, 945 26th ave so. 

Livingston, Frederick N, Ivellog Mill Co, Ballard. 

Loader, Daniel G, S & P S Packing Co. 

Lorig, Edward P, G N Elevator Co. 

Louy, Peter, 609 Cherry. 

Lawe Philip M, Ballard. 

Ludgate, Wallace G, Seattle Lumber Co. 

Lupky, Paul IP, Colfax & Rainier Sta., S Seattle. 

McDaniel, Wm E, P C B Co. 

McEachran, Jno, 225 Wall. 

Mclntire, G F, Milles & Geske. 

McLaughlin, Edward, 414 Jefferson. 

McLaughlin, Frank E, 425 Seneca. 

McLennon, Angus, Roy & Roy Mill Co. 

McMann, Jno, 1416 Eighth ave S. 

McMillan, Malcolm F. 1041% Main. 

MacDonald, David A, Fourteenth ave W, near Power House. 
Maddocks, Chas W, IPadfield & Roberts. 

Major, Chas W, 516% Maynard ave. 

Malcolm, Albert E. 518 Stewart. 

Manning, Abram, 2308 Third ave. 

Manley, Henry L, P C Co. 

Mann, Luther B, Wash Iron Works. 

Martin, Stephen T, 225 Ninth ave N. 

Matheson, Jas, 521 First ave S. 

Matthews, Wm B, Seattle Ice Co. 

Menhennick, J Lewis, Hotel Leland. 

Merriam, Ed F, Madison St Sta, S E Co. 

Miles, J O, S M Co. 

Miller, Albert, 423 Marion. 

Miller, Jno C, Pacific Lumber & Pipe Co. 

Mills, Herbert, 2703% First ave. 

Monahan, John G N, Elevator Co. 

Montgomery, Alexander H, Kimball ave, Green Lake. 


engineers’ manual. 


Moog, Chas F, James St Sta, S E Co. 

Moore, F B, 913 First aye. 

Moore, Frederick S, Latona Shingle Mills. 

Moore, Nathaniel D, P C Co. 

Moran, Jas, Fremont ave, Brooklyn. 

Moran, Jno W, 403% Pike. 

Moulton, Walter W, 68 Broad. 

Moyer, Wm H, 118 Broadway. 

Mulholland, Frank, Cascade Laundry. 

Murray, Robt, 1421 Fifth ave. 

Murray, Wm E, 911 Howell. 

Murphy, Thos H, Seattle Lumber Co. 

Nash, Jas, Green Lake. 

Nauer, Joseph, 1331 Twentieth ave S. 

Nelson, Geo, 2613% First ave. 

Newell, Isaac, Alki Point, W Seattle. 

Nichols, Horace, 2024 First ave. 

Niesen, Frank S, Wolf & Co. 

Nist, Jacob J, Q C Mfg Co. 

Nordquist, Axel T, 1711 Howard ave. 

Norwick, Chas, Van Asselt. 

Nowicki, Martin, Denion, Ballard. 

Nuber, Robert L, King & Grant, Latona. 

Nulter, Wm B, 2703% First ave. 

O’Brien, Wm, Grand Central Hotel. 

O’Brien, Wm F, 25th and Grand Boul, Interbay. 
Oliver, William E, 2127 Second ave. 

Olsen, Nils N, S E Co. 

Osgood, Clarence W P, 316 Twenty-seventh ave S. 
Osgood, Thaddeus, Taylor Laundry Co. 

Ostrand, Jno A, 1617 Bellevue ave. 

Page, Juvenile B, Sutherland Mill Co, Ltd. 

Palmer, Geo W, Zook Shingle Mill. 

Palmer, Warren P, Hotel Denver, Ballard. 

Parker, Geo F, S B & M Co. 

Parker, Isaac, 1120 Eighth ave. 

Parrott, Geo F, 2010 Western ave. 

Parsons, Alfred C, Newell M & M Co. 

Parsons, Roy, Newell M & M Co. 

Paulson, Paul, Rahlfs & Schoder. 

Pearson, Jno W, 1920 E Thomas. 

Peckham, A G. 823 First ave S. 

Pennington, Henry, Denny Clay Co. 

Peters, David M, 126 Wilbert, Ballard. 

Powell, Henry M, Powell Bros. 

Powers, Jno L, 512 First ave So. 

Price, S A, 1917 Terry ave. 

Pulsifer, Geo W, Madison Sta S E Co. 

Race, Chas F, N A T & T Co. 

Rand, Alfred, S M & N Co. 

Rebhahn, Jno A , S B & T Co. 

Reed, Sewell S, 1907 First ave. 

Rhines, Jos PI, If B Co. 

Richardson, Edward M, 512 Queen Anne ave. 

Riddell, Alexander C, 523 Prospect. 

Riley, Patrick J, cor Struria and Evanson, Fremont. 
Robertson, Jno, S Seattle. 

Robinson, Jos E, 919 Queen Anne Ave. 

Roe, Stephen C, Pacific and Colfax, S Seattle. 
Roope, Wm C, 123 W Mercer. 

Ross, Walter G, Seattle Fuel Co. 

Ross, Walter I, Seattle Fuel Co. 

Sadvall, Joseph, 420 Terrace. 

Scanlon, Joseph, P C Co. 

Schaefer, George, Hotel Butler. 

Severance, L, G, S E Co, new power house. 

Scheller, Jno P, S & P Mill Co. 

Schonfeldt, August, 2102 E Spruce. 



160 


ENGINEERS’ MANUAL. 


Schroeder, Herman C, Moran Bros Co. 

Seigert, Henry Carstens Packing Co. 

Sheedy, Jno, Novelty Mill Co. 

Sherry Frank, Duwatnisk. 

Simmons, Orson B, 1630 Thirteenth ave. 

Smart, Jno W, 900 Kilbourne, Fremont. 

Smith, Jas E, 705% Pike. 

Smith, Samuel W, 1006 E Thomas. 

Snow, Nelson L, 420 E Harrison. 

Snydpr, C W, 1514 Yesler way. 

Snydfy% M W, Rainier Grand Hotel. 

Stanton, Wm A, Western Mills. 

Stetson, Horatio A, 1143 12th ave S. 

Stoneman, C McL, Hotel Rainier. 

Strang, Chas F, Washington Cold Storage. 

Straw, Gilbert E, 337 Seventeenth ave N. 

Studdert, Hugh S, 505 Twenty-third ave. 

Sutherland, Jno B, 4301 Brooklyn ave, Brooklyn. 

Swartout, George V, Empire Laundry. 

Sylvester, Omar, 1826 Seventh ave. 

Yalley, Jno A, Coml Steam Boiler Works. 

Tanner, Richard I, Ninth ave S, S Seattle. 

Taylor, Edward F, P S B & D Co. 

Taylor, Wm, 7 Day, Ballard. 

Thompson, Benjamin, Chlopeck Fish Co. 

Thorn, Chas M, P S B & D Co. 

Tinkham, Edwin S, Yesler. 

Towner, Wm A, 104% Pike. 

Tribett, James T, 2306 Elliott ave. 

Turnbull, Jno D, 2610 Day. 

Tyndall, Henry J, Ballard Boiler Works. 

Valentine, Wm N, 133 Ballard ave, Ballard. 

Van Sehaick Augustus, Clise Investment Co. 

Van Vaikenburgh, Edgar W, 112 E Baker, Ballard. 

Van Winkle, Martin, 910 Twenty-First ave. 

Wade, Benjamin F, 2004 E Spruce. 

Wahl, Gus, Providence Hospital. 

AVallace, Fred H, 916 Jefferson. 

Wallace. Jno W, pumping sta, Ballard. 

Wamsley, Chas, 2931% First ave. 

Wardell, Arthur P, 514 Ninth ave. 

Warner, Fred, Seattle Lumber Co. 

Warwick, Geo, Broadway, Ballard. 

Watkins. Geo, Heinrich Bros Brewing Co. 

AVatson, Alfred, Richards, near Huron, Edgewater. 

Wheat, Joseph W, Rainier ave, near Stevens. 

AVhite, AValter P, South Park. 

AVhitford, James, 1905 First ave. 

Whittaker, Abraham, Moran Bros Co. 

Whittaker, James, S AA r & T Co. 

AVieck, Peter C, 734 Virginia. 

Wilhelm, Joseph J, Bellevue Hotel. 

Williams, Alfred J, 2210% Eighth ave. 

Williams, Arthur, H F Norton & Co. 

Williams, Benjamin, 413 Seventh ave S. 

AVilliams, James T, Q C Laundry. 

Williamson, Jno R, 2306 Second ave. 

Wills, Joseph I, 2403% Western ave. 

AVilson, Roy N, Twenty-third ave, Ilolgate. 

AVishard, Albert O, 501 Main. 

AVood, Frank E, City Pumping Station, Ballard. 

Wood, Geo AV, County Court House. 

Wood, Orville R, Novelty Mill Co. 

Wood, Robert, 1521 Third ave. 

Worle, Harry M, 225 Broadway. 

Wright, Hulet Q, Mutual Life Building. 

Wright, Jno C, Fourteenth ave W and Grand Boul, Interbay. 
Young, Walter O, 515 Seneca. 


ENGINEERS’ MANUAL. 


161 


THE CANADIAN BANK 
OE COMMERCE 

With which is amalgamated 

THE BANK OF BRITISH COLUMBIA. 

Head Office, TORONTO, ONT. Established 1867. 


Capital paid up $8,000,000.00 

(Eight Million Dollars) 

Surplus.$2,000,000.00 

Assets May 31, 1901 . . $67,553,578.13 


Accounts of Banks, Corporations, Firms and Individuals 

Solicited. 


Drafts issued available in any part of the World. 


Interest allowed on Time Deposits. 


Having established Branches at DAWSON, WHITE HORSE, 
SKAGWAY and ATLIN, this Bank has exceptional 
facilities for handling YUKON and 
ALASKA business. 


A General Banking Business Transacted. 


SEATTLE BRANCH : D. A. CAMERON, 

t, ' 

Cor. Second Ave. and James St. Manager. 









162 


ENGINEERS’ MANUAL 


THE DR. ALICE BEVERLEIGH 
PATENT 

Piston and Pump Packing 

Manufactured by 

BEVERLEIGH MEG. CO., Inc. 


This packing was invented and patented by Dr. Beverleigh, 
of London, England, who is a thorough, practical Chemist. 
After experimenting for years, the true combination of chemi¬ 
cals was discovered whereby a good water packing .was produc¬ 
ed that excels anything ever placed on the markets of the 
world. 

Dr. Beverleigh does not hesitate to challenge any inventor 
of water packing as regards the staying qualities of her inven¬ 
tion. This Packing can be utilized and adjusted to any shape 
required. Knowing that a good water packing is always in de¬ 
mand, Dr. Beverleight is fully prepared to meet it. 

Testimonials on application will show how it is appreciat¬ 
ed by reputable Engineers who have thoroughly tested it. 


BEVERLEIGH METAL POLISH 

has no equal, being free from all acids. Pre¬ 
vents rust and corrosion. Made by 


BEVERLEIGH MFG. CO 










































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